Torque and Work AP Physics C Mechanics FRQ – Exam Style Questions etc.
Torque and Work AP Physics C Mechanics FRQ
Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Question
A cylinder of mass M and radius R is released from rest at the top of an inclined plane, as shown above. The cylinder rolls without slipping down the incline. The rotational inertia of the cylinder is \(MR ^2/2\) .
(a) Derive an expression for the angular momentum of the cylinder about its center of mass when it has rolled a vertical distance h. Express your answer in terms of M, R, h, and physical constants, as appropriate.
(b) Is the angular momentum of the cylinder conserved as the cylinder rolls down the vertical distance h ? Yes No Justify your answer.
A child’s toy is composed of 3 narrow cylinders attached around a common axis through their centers, as shown above. The central cylinder has a mass M and radius R. Each of the two outer cylinders has a mass M and radius 2R.
(c) Derive an expression for the rotational inertia of the toy around its center. Express your answer in terms of M, R, and physical constants, as appropriate.
The toy is placed on a narrow track that is inclined at an angle of θ above the horizontal. Only the central cylinder is in contact with the track, as shown above. The toy rolls without slipping down the track.
(d)
i. On the dot below, which represents the toy, draw and label the forces (not components) that act on the toy. The dashed line is parallel to the narrow track. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
ii. Derive an expression for the linear acceleration of the toy as it rolls down the track. Express your answer in terms of M, R, θ , and physical constants, as appropriate.
iii. Determine the force of friction exerted on the toy as it rolls down the track. Express your answer in terms of M, R, θ , and physical constants, as appropriate.
(e) The toy reaches the bottom of the incline and rolls at constant speed v along a horizontal section of the track, as shown above. Derive an expression for the total kinetic energy of the toy while it is rolling. Express your answer in terms of M, R, v, and physical constants, as appropriate.
Answer/Explanation
(a) For correctly using conservation of energy\( U_{g1}+K_1=U_{g2}+K_2 U_{g1}+0=0+K_2\)
\(mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2\)
For correctly substituting into equation above for both linear and rotational kinetic energy
\(Mgh=\frac{1}{2}M(R\omega )^2+\frac{1}{2}\left ( \frac{MR^2}{2} \right )\omega ^2 gh=\frac{3}{4}R^2\omega ^2 \omega =\sqrt{\frac{4gh}{3R^2}}\)
For correctly substituting into the equation for angular momentum \(L=I\omega =\left ( \frac{MR^2}{2} \right )\left ( \sqrt{\frac{4gh}{3R^2}} =MR\sqrt{\frac{gh}{3}} \right )\)
(b) Selecting “No” For a correct justification Example: The frictional force exerts an external torque on the cylinder therefore the angular momentum of the cylinder is not conserevd. Example: The angular velocity is changing with h; therefore, L is changing.
(c) For setting the total rotational inertia for the toy equal to the sum of the rotaional inertias of the three cylinders \(T_tot=2I_{OUTER}+I_{INNER}=2\frac{MR^2}{2}_{OUTER}+\left ( \frac{MR^2}{2}_{INNER} \right )
I_{tot}=2\left ( \frac{M(2R)^2}{2} \right )+\left ( \frac{MR^2}{2} \right )=4MR^2+\frac{MR^2}{2}
I_{tot}=\frac{9}{2}MR^2\)
(d)i) For correctly drawing a vector representing the normal force perpendicular to the dashed line For correctly drawing a vector representing the weight of the toy directed straight down For correctly drawing a vector representing the frictional force directed up the incline Note: A maximum of two points can be earned if there are any extraneous vectors
ii) For correctly using Newton’s second law in linear form for the toy on the incline
\(F=ma\therefore (3M)g(sin\theta)-F_f=(3M)a
\varsigma =Ia\therefore F_fR=\left ( \frac{9}{2}MR^2 \right )\left ( \frac{a}{R} \right )\therefore F_f=\frac{9}{2}Ma
3Mg(sin\theta)-\frac{9}{2}Ma=3Ma\therefore 3g(sin\theta)=\frac{15}{2}a
a=\frac{2}{5}g(sin\theta)\)
(d) For an answer consistent with part (d)(ii)
iii) \(F_f=\frac{9}{2}M\left ( \frac{2}{5}g(sin\theta) \right )=\frac{9}{5}Mg(sin\theta)\)
(e) For using both rotational and linear kinetic energy for the total kinetic energy of the toy
\(K=K_L+K_R=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2\)
For correctly substituting for both the rotational inertia, the mass, and \(\omega =v/R\) into the equation above
\(K=\frac{1}{2}(3M)v^2+\frac{1}{2}\left ( \frac{9}{2}MR^2 \right )\omega ^2=\frac{3}{2}Mv^2+\frac{9}{4}MR^2\left ( \frac{v}{R} \right )^2=\frac{15}{4}Mv^2\)