Rotational, Circular, and Harmonic Motion - MCAT Chemical and Physical Foundations of Biological Systems

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Question

An amusement ride is used to teach students about centripetal force. The ride is a circlular wall that you place your back on. The wall and floor then begin to spin. Once it reaches a certain rotational velocity, the floor drops, and the students are pinned to the wall as a result of centripital force.

The diameter of the ride is 10m. The minimum frequency of rotation that results in students being pinned to the wall after the floor drops is 0.5Hz. What is the coefficient of static friction between the students and the wall?

$g= 10\frac{m}{s^2}$

Answer

There is quite a lot going on in this problem. However, we will take it one step at a time so that you can understand the logical progression of thought it takes to solve this problem.

The first major step will be converting the frequency into a centripetal acceleration.

Convert the frequency into a velocity:

$\text{velocity} = \text{circumference}\cdot \text{frequency}$

$v= \pi df = \pi(10m)(0.5Hz) = 5\pi \frac{m}{s}$

This is how fast the outer walls of the ride are traveling. We can convert this into centripetal acceleraton using the expression:

$a_c=\frac{v^2}{r}$

$a_c = \frac{\left ( 5\pi\frac{m}{s}\right )^2}{5m} = 5\pi^2\frac{m}{s^2}$

Then, we can create an expression for centripetal force for a student with mass :

$F_c = ma_c = m5\pi^2$

Now that we have a term for the centirpetal force, we can work toward finding the force of friction. The centripetal force is also the normal force used in calculating frictional force. If the student is pinned to the wall, then the frictional force is exactly equal to the student's weight.

$F_g = mg = F_{friction} = \mu_sN$

$mg = \mu_sN$

Plugging in our expression for the normal force:

$mg = \mu_s\left (m5\pi^2 \right )$

Cancel out mass and rearrange for the coefficient of friction:

$\mu_s = \frac{g}{5\pi^2} = \frac{10\frac{m}{s^2}}{5\pi^2\frac{m}{s^2}}$

$\mu_s = 0.20$

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Question

Which of these is the correct expression for torque, τ, when θ is the angle at which the force, F, acts around the axis of rotation?

τ = F cos θ

τ = F sin θ

τ = F tan θ

τ = F cotan θ

τ = F sec θ

Answer

Choice 2 is correct. Think about trying to change a tire…you apply the force on the tire iron as close to perpendicular as possible in order to generate the most force.

Since the sine of 90 degrees is 1, then you are applying the maximum torque you can generate according to the equation τ = F sin θ.

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Question

Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located at the 60cm mark with a weight of 30kg. At what point in between the two masses must the string be attached in order to balance the system?

Answer

This problem deals with torque and equilibrium. Noting that the string is between the two masses we can use the torque equation of $\tau_{ccw} = \tau _{cw}$ . We can use the equation $\tau = rFsin\theta$ to find the torque. Since force is perpendicular to the distance we can use the equation $\tau = rF$ (sine of 90o is 1). Force presented in this situation is gravity, therefore F=mg, and using the variable x as a placement for the string we can find r.

$\tau_{ccw} = \tau _{cw}$

$r_{1} (M_1) = r_{2}(M_2)$

x=43, thus the string is placed at the 43cm mark.

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Question

A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from it's left end. If a force of 50N is applied on it's right end, how much force would needs to be applied to the left end?

Answer

This a an example of rotational equilibrium involving torque. The formula for torque is $\tau = Fdsin\left ( \Theta \right )$ , where $\Theta$ is the angle that the force vector makes with the object in equilibrium and is the distance from the fulcrum to the point of the force vector. To achieve equilibrium, our torques must be equal.

$F_1d_1sin(\Theta_1)=F_2d_2sin(\Theta_2)$

Since the forces are applied perpendicular to the beam, $sin(\Theta)$ becomes 1. The distance of the fulcrum from the left end is 1m and its distance from the right end is 2m.

$\frac{(50N)(2m)}{1m}=100N=x$

Since the 50N force is twice as far from the fulcrum as the force that must be applied on the left side, it must be half as strong as the force on the left. The force on the left can be found to be 100N.

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Question

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod?

Answer

The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. Taking the fulcrum as the pivot point, the counterclockwise torque is due to the rod’s weight, gravitational force acting downwards at the center of the rod. If we use the pivot as our reference, then the center of the rod is 15cm from the reference.

$\small \small \small \tau_{ccw}=rF=(0.15m)(10\frac{m}{s^{2}})(0.2kg)=0.3Nm$

Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.

$\small \tau_{cw}=rF=r(10\frac{m}{s^{2}})(0.6kg)=0.3Nm$ .

Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the torque on the pulley when the system is motionless?

Answer

The net torque on the pulley is zero. Remember that $\tau=Fr$ , assuming the force acts perpendicular to the radius. Because the pulley is symmetrical in this problem (meaning the r is the same) and the tension throughout the entire rope is the same (meaning F is the same), we know that the counterclockwise torque cancels out the clockwise torque, thus, the net torque is zero.

In the image below, T1 (due to the platform with the 4 0.5kg weights) = T2 (the 2kg mass).

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

The rope is cut just above the 2kg mass, and the platform starts to fall. The rope is still wrapped around the pulley. What is the torque around the pulley?

Answer

In the instant the rope is cut right above the 2kg mass, the pulley now has net torque in the counterclockwise direction because the platform with the masses is still connected to the rope that winds around the pulley. Knowing that $\tau=Fr$ , we can determine the torque by knowing that the force (F) on the pulley is due to the tension (T) by the platform. The tension is 19.6N (equal to the weight) and knowing that the radius of the pulley is 0.25m, we can solve for torque.

$\tau=Fr=(19.6N)(0.25m)=4.9N*m$

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Question

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.

What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?

Answer

Torque is defined by the equation $\tau = Frsin\Theta$ . Since both students will exert a downward force perpendicular to the length of the seesaw, $sin\Theta=1$ . In our case, force is the force of gravity, given below, and is the distance from the center of the seesaw.

$F=mg=m(10\frac{m}{s^2})$

Since the torque must be zero in order for the seesaw to stay parallel (not move), the lighter student on the right must make his torque on the right equal to the torque of the student on the left. We can determine the required distance by setting their torques equal to each other.

$F_{1}r_{1} = F_{2}r_{2}$

$(60kg)(10\frac{m}{s^{2}})(3m) = (45kg)(10\frac{m}{s^{2}})r_{2}$

$r_{2} = 4 m$

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Question

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.

Imagine that the two students are sitting on the seesaw so that the torque is . Which of the following changes will alter the torque of the seesaw?

Answer

Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation. As a result, both students moving forward by one meter will cause a nonzero torque on the seesaw. This is because the heavier student's ratio of force and distance will result in less torque on his side than the lighter student.

$\tau=Fd$

$(60)g(d_1-1)\neq (45)g(d_2-1)$

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Question

One side of a seesaw carries a $\small 21kg$ mass four meters from the fulcrum and a $\smll 25.5kg$ mass two meters from the fulcrum. To balance the seesaw, what mass should be placed nine meters from the fulcrum on the side opposite the first two masses?

Answer

For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides.

$\tau=Fd=mgd$

The total torque must be equal on both sides in order for the net torque to be zero.

$\tau_1+\tau_2=\tau_3$

Substitute the formula for torque into this equation.

Now we can use the given values to solve for the missing mass.

The acceleration form gravity cancels from each term.

$84kg\cdot m+51kg\cdot m=m_3(9m)$

$135kg\cdot m=m_3(9m)$

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Question

An attraction at a science museum helps teach students about the power of torque. There is a long metal beam that has one pivot point. At one end of the bar hangs a full sedan, and on the other end is a rope at which students can pull down, raising the car off the ground.

The beam is 40 meters long and the pivot point is 5 meters from one end. A car of mass 500kg hangs from the short end of the beam. Neglecting the mass of the beam, what is the minimum mass of a student who can hang from the rope and begin to raise the car off the ground?

$g=10\frac{m}{s^2}$

Answer

We are trying to find what force needs to be applied to the rope to result in a net of zero torque on the beam.

Torque applied by the car:

$T = mgd = (500kg)(10\frac{m}{s^2})(5m) = 25,000 J$

We can use this to find the mass of a student that will create the same amount of torque while hanging from the rope:

$25,000 J = mgd = m(10\frac{m}{s^2})(35m)$

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Question

An insect sits at the edge of slowly turning wheel. The wheel is accelerated gradually until the insect can no longer hold on. The insect’s path of travel will be:

The insect will move away from the wheel in a straight line perpendicular to the wheel edge.

The insect will move away from the wheel in a straight line tangent to the wheel edge.

The insect will move away from the wheel edge in a curvilinear path.

The path is not predictable by Newtonian physics.

None of these is true.

Answer

Choice 2 is correct; this is how a classical slingshot operates. A body in motion tends to remain in motion unless a force disturbs that motion. At any point in time, the insect is moving in a straight line along a line tangent to the circumference of the wheel. When the little creature leaves the wheel, there is no longer any force acting on it, so it will move in a straight line, not a curvilinear path. Newtonian kinetics of course includes these concepts, because they relate closely to the motion of celestial bodies.

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Question

For an object traveling in a circle at a constant velocity of 100m/s, which of the following is true?

Answer

Since the object is traveling in a circle, it is constantly changing direction. This means that the direction of the velocity is also changing, even if the magnitude is not. Because the magnitude of the velocity does not change, we can assume that acceleration is constant.

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the linear velocity of the pulley the instant the rope to the 2kg mass is cut?

Answer

Note that we are asked the linear velocity of the pulley, not the angular velocity. With Newton’s second law, we know that F = ma. Additionally, in rotational motion, we know that $F=\frac{mv^2}{r}$ . The force (F) the instant the rope is cut is due to the platform with the 4 weights. Knowing the mass and radius of the pulley, we can solve for the linear velocity:

$F=\frac{mv^2}{r}$

$v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(19.6N)(0.25m)}{0.01kg}}=22.1m/s$

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Question

A ball with a mass of 0.95kg is attached to a string with a length of 75cm and moves in a vertical circle at a constant speed of 25m/s. What is the maximum tension in the string?

Answer

Maximum tension occurs when the ball is at its bottommost point, as the string experiences both the downward force of gravity and centripetal force.

$F_{c} = \frac{mv^{2}}{r}$

$F_{g} = mg$

$F_{net} = F_{c } + F_{g}$

$F_{net} = \frac{mv^{2}}{r} + mg = \frac{(0.95 kg)(25 m/s)^{2}}{0.75 m} + (0.95 kg)(10 m/s^{2}) = 800 N$

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Question

Two identical cars, A and B, are traveling on identical surfaces. Car A drives at a constant velocity in a perfectly circular path. Car B drives at a constant velocity along a long, straight path. Which best describes the accelerations, $\small a_{A}$ and $\small a_{B}$ , of the two cars?

Answer

An object moving in constant circular motion must have centripetal acceleration, directed towards the center of its path, in order to maintain its circular motion. So $\small a_{A}\neq0$ and is towards the center of car A's path.

An object moving linearly at constant velocity has 0 acceleration, so $\small a_{B}=0$ .

In general, we can think of acceleration as a change in either the magnitude or direction of an object's velocity. Neither of these cars has a change in the magnitude of its velocity, but car A does experience a change in the direction of its velocity, while car B does not.

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Question

What is the approximate centripetal acceleration of a point on the rim of a bicycle wheel with a diameter of $\small 60cm$ if the wheel is rotating at eight revolutions per second?

Answer

Centripetal acceleration is given by the equation:

$a_c=\frac{v^2}{r}$

We are given the angular velocity and the diameter. We can easily solve for the radius from the diameter, and use the angular velocity to find the linear velocity.

Convert the angular velocity to the proper units by changing revolutions to radians.

$\omega=\frac{8rev}{s}\frac{2\pi rad}{1rev}=16\pi\frac{rad}{s}$

Now we can convert the angular velocity to the linear velocity using the radius.

$v=\omegar=16\pi\frac{rad}{s}*0.30m=4.8\pi\frac{m}{s}$

Use the radius and the linear velocity to calculate the centripetal acceleration from the original equation.

$a_c=\frac{v^2}{r}=\frac{(4.8\pi\frac{m}{s})^2}{0.30m}$

$a_c=\frac{23.04\pi^2\frac{m^2}{s^2}}{0.30m}$

$a_c=76.8\pi^2\frac{m}{s^2}$

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Question

What is the centripetal force on a $\small 10kg$ ball being swung in a vertical loop with a velocity of $\small 5\frac{m}{s}$ on a $\small 1m$ string?

Answer

Centripetal force is given by the equation:

$F_c = ma_c=m\frac{v^2}{r}$

The length of the string represents the radius of the circle being formed. Use the given string length as the radius, along with the mass of the ball and the velocity, to calculate the centripetal force.

$F_c=(10kg)\frac{(5\frac{m}{s})^2}{1m}$

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Question

A man of mass is riding a roller coaster, which has a total mass of . As the ride enters a circular loop, it is traveling at a rate of $20 \frac{m}{s}$ . The loop has a radius of . If the average frictional force on the coaster is , how many g's of force does the man experience at the top of the loop?

$g = 10 \frac{m}{s^2}$

Answer

To calculate g's, we need to know the centripetal force the rider is experiencing. To calculate centripetal force, we need to know how fast the ride is traveling at the top of the loop. To do this, we will use the equation for conservation of energy:

If we set the height of the top of the loop to 10m, we can say that there is no initial potential energy. Since we are solving for final velocity, let's rearrange this for final kinetic energy:

Plugging in the expressions for each of these terms, we get:

$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2-mgh_f - F_fd$

Rearranging for final velocity:

$v_f = \sqrt{v_i^2-2gh_f-\frac{2F_fd}{m}}$

We have all of the values except for how far the coaster travels. It travels from the bottom to the top of the loop, which is half the circumference of the circle:

$d = \pi r = 5\pi$

Now we can plug in our values and solve:

$v_f = \sqrt{(20\frac{m}{s})^2-(2\cdot10\frac{m}{s^2}\cdot10m)-\frac{2\cdot600N\cdot5\pi}{1075kg}}$

$v_f = \sqrt{400\frac{m^2}{s^2} - 200\frac{m^2}{s^2} - 5.58\pi\frac{m^2}{s^2}} = \sqrt{182.47\frac{m^2}{s^2}} = 13.51 \frac{m}{s}$

Now that we know how fast the ride is traveling at the top of the loop, we can calculate the centripetal force on the man:

$F_c = ma_c = m\frac{v^2}{r}$

$F_c = (75kg)\frac{(13.51\frac{m}{s})^2}{5m} = 2737 N$

Now that we know the force in Newtons, we can convert this value to g's.

For this man:

$1'g' = mg = (75kg)(10\frac{m}{s^2}) = 750 N$

Therefore,

$G = \frac{2737N}{750\frac{N}{g}} = 3.65g$

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Question

A spring is compressed as far possible and is not permitted to expand. What can be said about its potential energy and its kinetic energy?

Answer

In this case, the kinetic energy of the spring is at a minimum. This is because, as the question indicates, the spring is not moving. At the same time, because the spring is compressed as far is it can be compressed, we know that its potential energy is at a maximum. The total energy of the spring therefore cannot be zero.

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