The rope shown in the figure is wound around a uniform cylinder of mass 4 kg , about the axis of the cylinder. If the cylinder rolls without slipping, then the linear acceleration of its centre of mass is

Question

The rope shown in the figure is wound around a uniform cylinder of mass 4 kg , about the axis of the cylinder. If the cylinder rolls without slipping, then the linear acceleration of its centre of mass is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-uoclavwihxdq.png" /> (A) 6.7 m s- 2 (B) 10.0 m s- 2 (C) 9.0 m s- 2 (D) none of these

Tardigrade Admin · Accepted Answer

For rotational motion τ=Iα ⇒ F(2 R)=(3/2)MR2((a/R)) ⇒ a=(4 F/3 M)=(4 (20)/3 (4))=6.7 m s- 2

Q. The rope shown in the figure is wound around a uniform cylinder of mass $4 k g$ , about the axis of the cylinder. If the cylinder rolls without slipping, then the linear acceleration of its centre of mass is

$6.7 m s^{- 2}$

$10.0 m s^{- 2}$

$9.0 m s^{- 2}$

none of these

For rotational motion
$τ = I α$
$\Rightarrow F (2 R) = \frac{3}{2} M R^{2} (\frac{a}{R})$
$\Rightarrow a = \frac{4 F}{3 M} = \frac{4 ( 20 )}{3 ( 4 )} = 6.7 m s^{- 2}$

Q. The rope shown in the figure is wound around a uniform cylinder of mass 4kg , about the axis of the cylinder. If the cylinder rolls without slipping, then the linear acceleration of its centre of mass is

6.7ms−2

10.0ms−2

9.0ms−2