Rotational Motion

JEE Main level, numerical answer, medium

153 of 280

Question 153 of 280

A light rope is wound around a hollow cylinder of mass 5kg5kg and radius 70cm70cm. The rope is pulled with a force of 52.5N52.5N. The angular acceleration of the cylinder will be _____ rads−2rad{s}^{-2}.

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Answer: 15

The formula to calculate the moment of inertia (I)\left(I\right) of the cylinder about its radial axis is given by
I=Mr2...(1)I=M{r}^{2}...\left(1\right)
Also, the torque (τ)\left(\tau \right) on the cylinder about the central axis because of the application of the external force is given by
τ=Fr...(2)\tau =Fr...\left(2\right)
Also, the torque can be expressed as
τ=Iα...(3)\tau =I\alpha ...\left(3\right)
Substitute the expressions from equation (1) and (2) into equation (3) and simplify to obtain the angular acceleration.
Fr=Mr2α⇒α=FMr...(4)\begin{matrix}Fr & = & M{r}^{2}\alpha \\ & \Rightarrow & \alpha =\dfrac{F}{Mr}...\left(4\right)\end{matrix}
Substitute the values of the known parameters into equation (4) to calculate the required angular acceleration.
α=52.5N5kg×0.70m=15rads−2\begin{matrix}\alpha & = & \dfrac{52.5N}{5kg\times 0.70m} \\ & = & 15rad{s}^{-2}\end{matrix}

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