Newton’s Second Law in Rotational Form AP Physics C Mechanics MCQ – Exam Style Questions etc.
Newton’s Second Law in Rotational Form AP Physics C Mechanics MCQ
Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
Question
Two blocks are joined by a light string that passes over the pulley shown above, which has radius R and moment of inertia I about its center. T1 and T2 are the tensions in the string on either side of the pulley and α is the angular acceleration of the pulley. Which of the following equations best describes the pulley’s rotational motion during the time the blocks accelerate?
(A) m2gR = Iα (B) T2R = Iα (C) (T2 – T1)R = Iα (D) (m2 – m1)gR = Iα
Answer/Explanation
Ans:C
\(\sum \)τ = T2R – T1R = Iα
Questions
A solid cylinder of mass m and radius R has a string wound around it. A person holding the string pulls it vertically upward, as shown above, such that the cylinder is suspended in midair for a brief time interval Δt and its center of mass does not move. The tension in the string is T, and the rotational inertia of the cylinder about its axis is \(\frac{1}{2}MR^{2}\)
Question(a)
the net force on the cylinder during the time interval Δt is
(A) mg (B) T – mgR (C) mgR – Τ (D) zero
Answer/Explanation
Ans:D
Solution: If the cylinder is “suspended in mid air” (i.e. the linear acceleration is zero) then \(\sum f = 0\)
Question(b)
The linear acceleration of the person’s hand during the time interval Δt is
(A) \(\frac{Tmg}{m}\) (B) 2g (C) \(\frac{g}{2}\) (D) \(\frac{T}{m}\)
Answer/Explanation
Ans:B
Solution: \(\sum \) τ = TR = Iα = ½ MR2α which gives α = 2T/MR and since \(\sum \)F = 0 then T = Mg so α = 2g/R the acceleration of the person’s hand is equal to the linear acceleration of the string around the rim of the cylinder a = αR = 2g
Question
A block of mass m is placed against the inner wall of a hollow cylinder of radius R that rotates about a vertical axis with a constant angular velocity ω, as shown above. In order for friction to prevent the mass from sliding down the wall, the coefficient of static friction μ between the mass and the wall must satisfy which of the following inequalities?
(A) \(\mu ^{}\frac{g}{^{2}R}\) (B) \(\mu ^{}\frac{^{2}R}{g}\) (C) \(\mu ^{}\frac{g}{^{2}R}\) (D) \(\mu ^{}\frac{^{2}R}{g}\)
Answer/Explanation
Ans:A
Solution: In order that the mass not slide down f = μFN ≥ mg and FN = mω2R
solving for μ gives μ ≥ g/ω2R