AP Physics 1 - Circular and Rotational Motion

A satellite is rotating once per minute. It has an moment of inertia of $10,000 kg \cdot m^2$ . Erin, an astronaut, extends the satellite's solar panels, increasing its moment of inertia to $30,000 kg \cdot m^2$ . How quickly is the satellite now rotating?

1 rotation every 3 minutes

3 rotations per minute

1 rotation every 9 minutes

9 rotations per minute

1 rotation per minute

Explanation

The formula for angular momentum is

$L=I\omega$

where

= angular momentum

= moment of inertia

$\omega$ = angular speed

Angular speed is defined as

$\omega = \frac{2\pi}{T}$

The initial period of the satellite is 1 minute, so:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{60s} = 0.105s^{-1}$

Plugging this in, we can solve for the initial angular momentum:

$L=(10000 kg \cdot m^2)(0.105s^{-1}) = 1050\frac{kg \cdot m^2}{s}$

After Erin extends the the solar panels, momentum is the same (conservation of momentum), but the moment of inertia is now $30,000 kg \cdot m^2$ . Thus,

$1050=(30000 kg \cdot m^2)\omega$

Therefore, $\omega = 0.035$

Plugging this back into the definition, we get

$\omega = \frac{2\pi}{T}$

$0.035 = \frac{2\pi}{T}$

Therefore,

Thus, the satellite now rotates once every 3 minutes.

1 rotation every 3 minutes

3 rotations per minute

1 rotation every 9 minutes

9 rotations per minute

1 rotation per minute

Explanation

The formula for angular momentum is

$L=I\omega$

where

= angular momentum

= moment of inertia

$\omega$ = angular speed

Angular speed is defined as

$\omega = \frac{2\pi}{T}$

The initial period of the satellite is 1 minute, so:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{60s} = 0.105s^{-1}$

Plugging this in, we can solve for the initial angular momentum:

$L=(10000 kg \cdot m^2)(0.105s^{-1}) = 1050\frac{kg \cdot m^2}{s}$

After Erin extends the the solar panels, momentum is the same (conservation of momentum), but the moment of inertia is now $30,000 kg \cdot m^2$ . Thus,

$1050=(30000 kg \cdot m^2)\omega$

Therefore, $\omega = 0.035$

Plugging this back into the definition, we get

$\omega = \frac{2\pi}{T}$

$0.035 = \frac{2\pi}{T}$

Therefore,

Thus, the satellite now rotates once every 3 minutes.

A child is swinging a toy rocket around their head with a rope. If they double the frequency of "orbits", how will the tension in the rope change?

Quadrupled

Doubled

Halved

Quartered

None of these

Explanation

The tension in the rope is equal to the centripetal force on the rocket.

$T=m\frac{v^2}{r}$

Definition of velocity:

$v=\frac{d}{t}$

The time necessary to go around one circle is it's period, which is the inverse of the frequency.

Doubling the frequency would double the velocity, which would quadruple the tension.

I start pushing a merry-go-round with a torque of 10 Newton-meters. It has a moment of inertia of $100 kg \cdot m^2$ . What is its rotational speed after 3 seconds assuming it starts at rest?

$\omega = 0.3s^{-1}$

$\omega = 30s^{-1}$

$\omega = 0.3s$

$\omega = 30s$

$\omega = 3s^{-1}$

Explanation

The angular moment of the merry-go-round after 3 seconds is simply

$L=\tau t=(10 N \cdot m)(3s)=30 \frac{kg \cdot m^2}{s}$

Angular momentum is also given by

$L=I\omega=(100 kg \cdot m^2)\omega$

Plugging in 30 for gives us

$\omega = 0.3s^{-1}$

A object moves in a circular motion with a diameter of . What is the magnitude of this object's velocity?

$5: \frac{m}{s}$

$4: \frac{m}{s}$

$6: \frac{m}{s}$

$3: \frac{m}{s}$

$2: \frac{m}{s}$

Explanation

Let's start with what we know, and what we want to know. We are given the mass of an object, and we're told that it's traveling in a circular orbit. Also, we have been given the diameter of the orbit. What we are trying to find is the velocity of the object.

In this problem, we'll need to consider the centripetal force of the object as it travels in a circular orbit. The force that is contributing to the centripetal force is the weight of the object. Therefore, we can write the following expression:

$F_{Centripetal}=F_{Weight}$

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

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

$v=\pm \sqrt{rg}$

The above expression tells us that the object's velocity is dependent on the acceleration due to gravity, as well as the radius of the orbit. Since we're told that the diameter of the orbit is , we can find the radius by taking half of this value, which is . Moreover, the plus-or-minus sign indicates that the object can either be traveling in a clockwise or counter-clockwise orbit. In either case, the object will have the same magnitude of velocity.

Solving for velocity, we can plug in values to obtain:

$v=\sqrt{rg}=\sqrt{(2.55: meters)(9.8: \frac{meters}{second^{2}})}=5:\frac{m}{s}$

Terry is pushing a vertical lever that is attached to the floor, and he pushes above the point of rotation. If he pushes with a force of at an angle of $60^{\text o}$ from the ground, what is the magnitude of torque that he is applying to the lever's hinge?

Explanation

Magnitude of torque can be found by relating the amount of force applied perpendicular to a lever arm about a point of rotation.

$\tau=rF$

In this case, the force is not perpendicular, so we must take the perpendicular aspect of the force to find torque.

$\tau=rFsin\theta$

Plug in and solve.

$\tau=1.5m(100N)sin(30^{\text o})$

$\tau=75Nm$

I start pushing a merry-go-round with a torque of 10 Newton-meters. It has a moment of inertia of $100 kg \cdot m^2$ . What is its rotational speed after 3 seconds assuming it starts at rest?

$\omega = 0.3s^{-1}$

$\omega = 30s^{-1}$

$\omega = 0.3s$

$\omega = 30s$

$\omega = 3s^{-1}$

Explanation

The angular moment of the merry-go-round after 3 seconds is simply

$L=\tau t=(10 N \cdot m)(3s)=30 \frac{kg \cdot m^2}{s}$

Angular momentum is also given by

$L=I\omega=(100 kg \cdot m^2)\omega$

Plugging in 30 for gives us

$\omega = 0.3s^{-1}$

A child is swinging a toy rocket around their head with a rope. If they double the frequency of "orbits", how will the tension in the rope change?

Quadrupled

Doubled

Halved

Quartered

None of these

Explanation

The tension in the rope is equal to the centripetal force on the rocket.

$T=m\frac{v^2}{r}$

Definition of velocity:

$v=\frac{d}{t}$

The time necessary to go around one circle is it's period, which is the inverse of the frequency.

Doubling the frequency would double the velocity, which would quadruple the tension.

Pluto distance to sun: $1.43*10^{9}km$

Determine the centripetal acceleration of Pluto with respect to the Sun.

$9.09*10^{-7}\frac{m}{s^2}$

$5.87*10^{-7}\frac{m}{s^2}$

$10.01*10^{-7}\frac{m}{s^2}$

$7.57*10^{-7}\frac{m}{s^2}$

$8.44*10^{-7}\frac{m}{s^2}$

Explanation

$a_{centri}=\frac{v^2}{r}$

$perimeter~of~circle:2\pi r$

$v=\frac{d}{t}$

Combine equations:

$a_{centri}=\frac{\frac{(2\pi r)^2}{t^2}}{r}$

Convert to meters and seconds and plug in values:

$a_{centri}=\frac{\frac{(2\pi 1.43*10^{12})^2}{(7.88*10^9)^2}}{1.43*10^{12}}$

$a_{centri}=9.09*10^{-7}\frac{m}{s^2}$

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure).

If the maximum centripetal force felt by mass A is , what is the minimum centripetal force it feels? Neglect air resistance and internal friction.

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

Explanation

First we need to determine when at which points mass A will be experiencing the most/least centripetal force. The most will be when the rod is vertical and mass A is at the bottom (initial condition). This is because the potential energy is minimized (thus kinetic is maximized) when the large mass is at its low point. Therefore, the lest centripetal force will be when mass A is at the top (final condition). So now let's begin with the equation for conservation of energy:

Unfortunately we can't eliminate any terms, so let's begin expanding them one at a time:

$U_i = mgh_i=m_Agh_{A_{i}}+m_Bgh_{B_{i}}$

If we assume that the lowest point of the circle has a height of 0, the initial potential energy of mass A will be 0:

$U_i = m_Bgh_{B_{i}}$

We also know that the height at the top of the circle is exactly one rod length from our reference height, so we can say:

Moving on to initial kinetic:

$K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}m_Av_{A_{i}}^2+ \frac{1}{2}m_Bv_{V_{i}}^2$

Since bot velocities are always the same, we can say:

$K_i = \frac{1}{2}v_i^2(m_A+m_B)$

Now final kinetic. We'll just skip ahead to it's final form:

Note how it's the same as before, except that the masses are switched. No final kinetic. We'll skip to it's final form again:

$K_f= \frac{1}{2}v_f^2(m_A+m_B)$

Now plugging all of these back into the expression for conservation of energy, we get:

$m_Bgl+ \frac{1}{2}v_i^2(m_A+m_B) = m_Agl+ \frac{1}{2}v_f^2(m_A+m_B)$

In order to find the final centripetal force, we are going to need to need the final velocity, so let's begin rearranging for that:

Factoring and dividing, we get:

$v_f^2=\frac{2gl(m_B-m_A)+v_i^2(m_A+m_B)}{(m_A+m_B)}=\frac{2gl(m_B-m_A)}{(m_A+m_B)}+v_i^2$

$v_f=\sqrt{\frac{2gl(m_B-m_A)}{(m_A+m_B)}+v_i^2}$

We know every variable except initial velocity, which we can solve for by using the equation for centripetal force:

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

Where mass is mass B, v is the initial velocity, and radius is half the length of the rod:

$F_c = m_A\frac{v_i^2}{\left ( \frac{1}{2}l\right )}$

Now we rearrange for initial velocity:

$F_c = 2m_A\frac{v_i^2}{l}$

$v_i=\sqrt{\left (\frac{F_cl}{2m_A} \right )} = \sqrt{\frac{(300N)(4m)}{2(10kg)}}$

$v_i = 7.75\frac{m}{s}$

Plugging this back into our big equation along with the rest of the variable:

$v_f=\sqrt{\frac{2(10\frac{m}{s^2})(4m)(5kg-10kg)}{(10kg+5kg)}+\left ( 7.75\frac{m}{s}^2\right )}$

$v_f = 5.77\frac{m}{s}$

Now using the expression for centripetal force to find our final answer:

$F_c = 2m_A\frac{v_f^2}{l}=(2)(10kg)\frac{\left ( 5.7\frac{m}{s}\right )^2}{(4m)}$

Circular and Rotational Motion

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