An inextensible thread is wound round a cylinder of mass $M=7.0 \mathrm{~kg}$ and radius $R=10 \mathrm{~cm}$. The cylinder is placed over a smooth horizontal surface and thread is passed over a massless and frictionless pulley such that the part of the thread between cylinder and pulley is…

Question

An inextensible thread is wound round a cylinder of mass $M=7.0 \mathrm{~kg}$ and radius $R=10 \mathrm{~cm}$. The cylinder is placed over a smooth horizontal surface and thread is passed over a massless and frictionless pulley such that the part of the thread between cylinder and pulley is horizontal as shown in figure.
A block of mass $m=1.0 \mathrm{~kg}$ is attached with free end of thread. If the system is released from rest, calculate acceleration of the block. $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$.

Aicharya · Accepted Answer

As horizontal surface is smooth, and force is acting on the e cylinder, therefore the cylinder will slip over the surface.
Let acceleration of axis of the cylinder be $a_{1}$ (rightward and its angular acceleration be $\alpha$ (clockwise) then downward acceleration of the block will be equal to,
$a_{2}=a_{1}+R \alpha$
Taking moment of forces acting on the cylinder about $0 .$
$\begin{aligned} T \cdot R &= I \alpha \text { where } I=\frac{M R^{2}}{2} \\ T &=\frac{1}{2} M R \alpha \quad \text{...(i)} \end{aligned}$ For horizontal forces on the cylinder,
$T=M a_{1}$ ...(ii)
From Eqs. (i) and (ii)
$R \alpha=2 a_{1}$
hence $a_{2}=3 a_{1}$
Now considering forces acting on the block,
$m g-T=m a_{2}$
Substituting $T=M a_{1}$ and $a_{2}=3 a_{1}$
$a_{1}=\frac{m g}{(M+3 m)} \text { or } a_{1}=1 \mathrm{~ms}^{-2}$
Hence, acceleration of the block, $a_{2}=3 \mathrm{~ms}^{-2}$

An inextensible thread is wound round a cylinder of mass \(M=7.0 \mathrm{~kg}\) and radius \(R=10…

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