AP EAPCET 2021 - 20th August Morning Shift Physics - Rotational Motion Question 38 English Explanation

Let, acceleration of body = a

Friction force = $$f$$

Mass of body = $$m$$

Radius of body = R

$$mg\sin\theta-f=ma$$ ..... (i)

Torque, $$\tau=Rf=I\alpha$$

$$\Rightarrow f=I\alpha/R$$ ..... (ii)

From Eqs. (i) and (ii), we get

$$m g \sin \theta-\frac{I \alpha}{R}=m a$$

Angular acceleration, $$\alpha=a / R$$

$$\begin{array}{ll} \Rightarrow & m g \sin \theta-\frac{I a}{R^2}=m a \\ \Rightarrow & m g \sin \theta=a\left(m+\frac{I}{R^2}\right) \\ \Rightarrow & a=\frac{m g \sin \theta}{m+\frac{I}{R^2}}=\frac{g \sin \theta}{1+\frac{I}{m R}} \end{array}$$

$$\because$$ Distance covered by body, $$s=\frac{1}{2} a t^2$$

$$\begin{aligned} & \because \quad s_{\text {sphere }}=s_{\text {cylinder }} \\ & \frac{1}{2} a_{\text {sphere }} \times t^2=\frac{1}{2} \times s_{\text {cylinder }} \times t^2 \\ & \Rightarrow \quad a_{\text {sphere }}=a_{\text {cylinder }} \\ & \Rightarrow \quad \frac{g \sin 30 \Upsilon}{1+\frac{\frac{2}{5} m R^2}{m R^2}}=\frac{g \sin \theta}{1+\frac{m R^2}{m R^2}} \\ & \Rightarrow \quad \frac{1 / 2}{7 / 5}=\frac{\sin \theta}{2} \Rightarrow \sin \theta=\frac{5}{7} \\ & \Rightarrow \quad \theta=\sin ^{-1}\left(\frac{5}{7}\right)=45.6 \Upsilon \approx 45 \Upsilon \\ & \end{aligned}$$

Rotational Motion

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