A cord is wound around the circumference of a bicycle wheel (without$$ $$tyre) of diameter $$1$$ m. A mass of $$2$$ kg is tied to the end of the cord and it is allowed to fall from rest. The weight falls $$2m$$ in $$4$$ s. The axle of the wheel is horizontal and the wheel rotates with its plane vertical. (Take$$ $$g=10ms$$−$$2$$ $$) then the angular acceleration of the wheel is (rad/s2)

Question

A cord is wound around the circumference of a bicycle wheel (without$$ $$tyre) of diameter $$1$$ m. A mass of $$2$$ kg is tied to the end of the cord and it is allowed to fall from rest. The weight falls $$2m$$ in $$4$$ s. The axle of the wheel is horizontal and the wheel rotates with its plane vertical. (Take$$ $$g=10ms$$−$$2$$ $$) then the angular acceleration of the wheel is (rad/s2)

Infinity Learn · Accepted Answer

We first find the linear acceleration a by using the relation

$\begin{array}{l} s = ut + \frac{1}{2} {at}^{2} \\ 2 = 0 + \frac{1}{2} \times a \times (4)^{2} \end{array}$

which gives $a = \frac{1}{4} {ms}^{- 2}$ . Now R = 0.5 m. The angular acceleration $α$ is

$α = \frac{a}{R} = \frac{1}{4 \times 0 .5} = \frac{1}{2} = 0 .5 {rads}^{- 2}$

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Rotational motion

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Rotational motion

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We first find the linear acceleration a by using the relation s=ut+12at22=0+12×a×(4)2which gives a=14ms−2. Now R = 0.5 m. The angular acceleration α isα=aR=14×0.5=12=0.5rads−2.

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