Newton's Second Law in Rotational Form
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AP Physics C: Mechanics › Newton's Second Law in Rotational Form
A wheel experiences net torque $\tau_{\text{net}}$; by $\tau=I\alpha$, how does increasing $\tau_{\text{net}}$ affect rotational motion if $I$ is constant?
It increases linear acceleration only
It decreases $\alpha$ proportionally
It increases $\alpha$ proportionally
It increases $I$ but not $\alpha$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. Newton's Second Law in rotational form is expressed as τ = Iα, indicating that net torque causes angular acceleration proportional to the moment of inertia. When the moment of inertia I is constant, the relationship shows that torque and angular acceleration are directly proportional. Choice B is correct because increasing τ_net while keeping I constant will proportionally increase α, as α = τ_net/I demonstrates a direct linear relationship. Choice A is incorrect because it suggests an inverse relationship, contradicting the fundamental equation. To help students: Draw parallels to F = ma in linear motion; use numerical examples showing how doubling torque doubles angular acceleration; emphasize that moment of inertia plays the same role as mass in resisting changes to rotational motion.
On a disc, torque satisfies $\tau=rF\sin\theta$ and $\tau=I\alpha$. If $F$ stays constant and tangential, what happens to $\tau$ when $r$ doubles?
$\tau$ stays constant
$\tau$ halves
$\tau$ becomes $r^2F$
$\tau$ doubles
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. Torque (τ) is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the lever arm and θ is the angle between force and lever arm. For a tangential force (θ = 90°), the torque equation simplifies to τ = rF since sin90° = 1. Choice A is correct because when r doubles while F remains constant and tangential, the torque τ = rF also doubles, demonstrating the direct proportionality between torque and lever arm. Choice B is incorrect because it suggests an inverse relationship, which would only occur if we were considering angular acceleration with changing moment of inertia. To help students: Use visual demonstrations showing how increasing the lever arm increases the rotational effect; emphasize that torque is directly proportional to both force and lever arm distance; practice problems where students identify which variables change and predict the effect on torque.
Torque is $$\tau = rF\sin\theta$$ and $$\tau_{\text{net}} = I\alpha$$ connects net torque to angular acceleration. A disc has $I=0.50,\text{kg\u00b7m}^2$ and two tangential forces act at the rim: $8,\text{N}$ clockwise and $3,\text{N}$ counterclockwise at $r=0.25,\text{m}$. What is $\alpha$ (clockwise positive)?
$-2.5,\text{rad/s}^2$
$5.5,\text{rad/s}^2$
$0.63,\text{rad/s}^2$
$2.5,\text{rad/s}^2$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form with multiple torques and sign conventions. When multiple torques act, the net torque is the algebraic sum considering direction: τ_net = τ_clockwise - τ_counterclockwise = (0.25 m)(8 N) - (0.25 m)(3 N) = 2.0 - 0.75 = 1.25 N·m clockwise. Applying Newton's Second Law τ = Iα gives the angular acceleration. Choice C is correct because α = τ_net/I = 1.25 N·m / 0.50 kg·m² = 2.5 rad/s² in the clockwise (positive) direction. Choice A is incorrect because it has the wrong sign, suggesting counterclockwise rotation when the net torque is clockwise. To help students: Practice problems with multiple torques; establish clear sign conventions before solving; use free-body diagrams adapted for rotation showing all torques and their directions.
Torque is $$\tau = rF\sin\theta$$ and rotational Newton’s Second Law is $$\tau_{\text{net}} = I\alpha$$. A disc experiences $\tau_{\text{net}}=9.0,\text{N\u00b7m}$ and $\alpha=3.0,\text{rad/s}^2$. Determine the moment of inertia $I$.
$0.33,\text{kg\u00b7m}^2$
$3.0,\text{kg\u00b7m}^2$
$27,\text{kg\u00b7m}^2$
$6.0,\text{N\u00b7m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and calculating moment of inertia from torque and angular acceleration. Newton's Second Law in rotational form τ = Iα can be rearranged to I = τ/α when solving for moment of inertia. This shows that moment of inertia represents an object's resistance to angular acceleration for a given torque. Choice B is correct because it accurately calculates: I = 9.0 N·m / 3.0 rad/s² = 3.0 kg·m². Choice A is incorrect because it multiplies τ and α instead of dividing, showing a fundamental misunderstanding of the equation's structure. To help students: Practice algebraic manipulation of τ = Iα for different unknowns; emphasize that I plays the same role as mass in F = ma; use unit analysis to check that kg·m² results from N·m divided by rad/s².
A disc follows $\tau=rF\sin\theta$ and $\tau=I\alpha$. If the same $F$ is applied at the same $r$ but changes from tangential to radial, what happens to $\alpha$?
$\alpha$ stays the same
$\alpha$ becomes $F/I$
$\alpha$ becomes zero
$\alpha$ doubles
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. The key concept is that torque depends on the perpendicular component of force, expressed through τ = rFsinθ where θ is the angle between the force vector and the lever arm. A radial force acts along the lever arm, making θ = 0°, while a tangential force acts perpendicular to it with θ = 90°. Choice A is correct because when the force becomes radial, θ = 0° and sin0° = 0, making τ = 0, which results in α = τ/I = 0. Choice B is incorrect because it ignores the angular dependence of torque, assuming the force direction doesn't matter. To help students: Use door-opening analogies to show why pushing radially at the hinge produces no rotation; demonstrate with physical objects how force direction affects rotational motion; emphasize that torque is a vector quantity with both magnitude and direction considerations.
Torque is $$\tau = rF\sin\theta$$ and rotational Newton’s Second Law is $$\tau_{\text{net}} = I\alpha$$, where $I$ resists changes in rotation. A wheel has $I=2.0,\text{kg\u00b7m}^2$ and experiences a net torque $\tau_{\text{net}}=6.0,\text{N\u00b7m}$. Calculate the angular acceleration $\alpha$.
$3.0,\text{N\u00b7m}$
$3.0,\text{rad/s}^2$
$12,\text{rad/s}^2$
$0.33,\text{rad/s}^2$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and calculating angular acceleration from given torque and moment of inertia. Newton's Second Law in rotational form is expressed as τ = Iα, where net torque causes angular acceleration proportional to the moment of inertia. Rearranging this equation gives α = τ/I, allowing us to find angular acceleration when torque and moment of inertia are known. Choice B is correct because it accurately applies this relationship: α = 6.0 N·m / 2.0 kg·m² = 3.0 rad/s². Choice D is incorrect because it gives the torque value with units of N·m rather than the requested angular acceleration in rad/s², showing confusion about what quantity to calculate. To help students: Practice identifying what quantity is being asked for; emphasize unit analysis to verify answers; create analogies between F = ma and τ = Iα to reinforce conceptual understanding.
A disc obeys $\tau=I\alpha$. If $\alpha=2.5,\text{rad/s}^2$ and $I=1.2,\text{kg·m}^2$, what net torque $\tau_{\text{net}}$ acts?
$3.0,\text{N·m}$
$4.8,\text{N·m}$
$0.48,\text{N·m}$
$3.0,\text{rad/s}^2$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. The equation τ = Iα directly relates net torque to the product of moment of inertia and angular acceleration. This fundamental relationship allows calculation of any quantity when the other two are known. Choice B is correct because τ_net = Iα = 1.2 × 2.5 = 3.0 N·m, demonstrating straightforward application of Newton's Second Law in rotational form. Choice D is incorrect because it likely multiplies by 2 somewhere in the calculation, possibly confusing this with a different formula. To help students: Create a reference sheet comparing F = ma and τ = Iα; practice direct substitution problems before moving to more complex scenarios; emphasize checking units to ensure the answer makes physical sense.
A wheel’s torque is $\tau=rF\sin\theta$ and also $\tau=I\alpha$. If $\alpha=4.0,\text{rad/s}^2$ and $I=0.75,\text{kg·m}^2$, what is $\tau_{\text{net}}$?
$3.0,\text{rad/s}^2$
$12,\text{N·m}$
$0.19,\text{N·m}$
$3.0,\text{N·m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. Newton's Second Law in rotational form states τ = Iα, where torque equals the product of moment of inertia and angular acceleration. This relationship allows us to calculate any of the three quantities when the other two are known. Choice B is correct because τ_net = Iα = 0.75 × 4.0 = 3.0 N·m, demonstrating proper application of the rotational form of Newton's Second Law. Choice C is incorrect because it has units of angular acceleration (rad/s²) rather than torque (N·m), showing confusion between the quantities in the equation. To help students: Practice identifying which quantity to solve for in τ = Iα; emphasize unit analysis to catch errors; create a comparison table showing linear (F = ma) and rotational (τ = Iα) forms of Newton's Second Law.
A disc has $\tau=rF\sin\theta$ and $\tau=I\alpha$. A $10,\text{N}$ force acts at $r=0.30,\text{m}$ at $\theta=30^\circ$; $I=0.50,\text{kg·m}^2$. Find $\alpha$.
$3.0,\text{rad/s}^2$
$6.0,\text{rad/s}^2$
$0.30,\text{rad/s}^2$
$1.0,\text{N·m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. This problem requires calculating torque using τ = rFsinθ with a non-perpendicular force, then applying τ = Iα to find angular acceleration. The angle θ is crucial because it determines the perpendicular component of the force that creates torque. Choice A is correct because τ = rFsinθ = 0.30 × 10 × sin30° = 0.30 × 10 × 0.5 = 1.5 N·m, then α = τ/I = 1.5/0.50 = 3.0 rad/s². Choice B is incorrect because it likely uses sin30° = 1 instead of 0.5, doubling the calculated torque and angular acceleration. To help students: Emphasize that only the perpendicular component of force creates torque; practice problems with various angles; use vector diagrams to visualize force components and identify the angle θ in the torque equation.
A disc’s torque is $\tau=rF\sin\theta$ and $\tau=I\alpha$. With $r=0.40,\text{m}$, $F=20,\text{N}$ tangential, and $I=2.0,\text{kg·m}^2$, calculate $\alpha$.
$8.0,\text{rad/s}^2$
$0.25,\text{rad/s}^2$
$4.0,\text{rad/s}^2$
$8.0,\text{N·m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. For a tangential force, the torque calculation simplifies because θ = 90° and sin90° = 1, making τ = rF. This torque then determines the angular acceleration through Newton's Second Law in rotational form. Choice B is correct because τ = rF = 0.40 × 20 = 8.0 N·m (since the force is tangential), then α = τ/I = 8.0/2.0 = 4.0 rad/s². Choice C is incorrect because it likely uses the torque value as the angular acceleration, confusing the two different quantities. To help students: Emphasize that tangential means perpendicular to the radius, simplifying calculations; practice identifying when sinθ = 1; use unit analysis to distinguish between torque (N·m) and angular acceleration (rad/s²).