Show that the angular acceleration $a$ of the cylinder is $\frac{g}{3R}$

Show that the tension T in the string is$\frac{Mg}{6}$

The block falls a distance 0.50 m after its release before hitting the ground. Show that the block hits the ground 0.55 s after release.

Calculate, for the cylinder, at the instant just before the block hits the ground the angular momentum.

Calculate, for the cylinder, at the instant just before the block hits the ground the kinetic energy.

equations of motion are: TR = $\frac{1}{2}$ MR² $a$ and $\frac{Mg}{4}$ − T = $\frac{M}{4}$ $a$

OR

$\frac{M}{4}$ gR = $\frac{1}{2}$ MR² $a$ + $\frac{M}{4}$ Ra

use of a = $a$R ✔

combine equations to get result ✔

Allow energy conservation use.

This is a show that question, so look for correct working.

Do not allow direct use of tension from a ii)

use of T = $\frac{1}{2}$MR$a$ to find T = $\frac{1}{2}$ MR × $\frac{g}{3R}$  ✔

«cancelling to show final answer»

a = 3.27 «ms⁻²» / a = g/3 ✔

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 0.50}{3.27}}$  ✔

= 0.55 «s»

Do not apply ECF from MP1 to MP2 if for a=g, giving answer 0.32 s.

ALTERNATIVE 1

ΔL «= ΓΔt = TRΔt » = $\frac{12\times 9.81\times 0.20\times 0.55}{6}$  ✔

ΔL = 2.2J «Js»✔

ALTERNATIVE 2

ω =<$a$Δt = $\frac{g}{3R}$ Δt = $\frac{9.81\times 0.55}{3\times 0.20}$ = > 8.99 «rads⁻¹» ✔

ΔL «=Iω»  $\frac{1}{2}$ × 12 × 0.20² × 8.99 = 2.2 «Js»

Award [2] for a bald correct answer.

ω =<$a$Δt = $\frac{g}{3R}$ Δt = $\frac{9.81\times 0.55}{3\times 0.20}$ = > 8.99 «rads⁻¹» ✔

E_k = «$\frac{1}{2}$ Iω² = $\frac{1}{4}$ MR² ω² = $\frac{1}{4}$ × 12 × 0.20² × 8.99² = » 9.7 «J» 

Award [2] for a bald correct answer.

This simple system of a solid cylinder rotating about a fixed axle and a block hanging on the rope accelerating proved to be rather difficult for many candidates. Only the best candidates were able to solve this problem, by writing two separate equations, and then solving to establish the acceleration of the system. The use of energy conservation was not a common technique attempted by students. As this is a “show that” question, candidates were asked to show that the angular acceleration is equal to g/(3R). Quite a high number of candidates successfully used the equations of motion and a similar number used the moment of inertia of the block around the axle. 

This simple system of a solid cylinder rotating about a fixed axle and a block hanging on the rope accelerating proved to be rather difficult for many candidates. Only the best candidates were able to solve this problem, by writing two separate equations, and then solving to establish the acceleration of the system. The use of energy conservation was not a common technique attempted by students. As this is a “show that” question, candidates were asked to show that the angular acceleration is equal to g/(3R). Quite a high number of candidates successfully used the equations of motion and a similar number used the moment of inertia of the block around the axle. In ii), well-prepared candidates used the angular acceleration of the cylinder from i) and the tension expression to arrive at the given Mg/6.

The application of the uniformly accelerated motion to this context was mastered successfully by better candidates.

Angular speed, angular momentum and the kinetic energy of the cylinder at the instant given were successfully calculated by only the better-prepared candidates. Students were mostly able to recognize the correct expressions to use but struggled to find the angular speed in many instances. Error carried forward was very common throughout this question for students who incorrectly determined the angular velocity but understood and correctly applied the momentum and energy equations.

Angular speed, angular momentum and the kinetic energy of the cylinder at the instant given were successfully calculated by only the better-prepared candidates. Students were mostly able to recognize the correct expressions to use but struggled to find the angular speed in many instances. Error carried forward was very common throughout this question for students who incorrectly determined the angular velocity but understood and correctly applied the momentum and energy equations. In ii), quite a high number of candidates misread the question and calculated the energy of the whole system.