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Question

A pendulum consists of a rod of mass 1 kg and length 1 m connected to a pivot with a solid sphere attached at the other end with mass 0.5 kg and radius 30 cm. What is the torque about the pivot when the pendulum makes an angle of 30° with respect to the vertical?

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First, we draw a representation of the pendulum in order for us to have a better idea on how to solve the exercise. Then, we continue to calculate the torque about the pivot point.

τ=Mrodg(l2)sin(θ)+Msphereg(l+R)sin(θ)=gsin(θ)[Mrodl2+Msphere(l+R)]=(9.8)sin(30°)[(1)(1)2+(0.5)(1+0.30)]=5.64 Nm\begin{align*} \tau&=M_\text{rod} \: g \left(\dfrac{l}{2}\right) \sin\left(\theta\right) + M_\text{sphere} \: g \left(l+R\right) \sin\left(\theta\right) \\ &=g \sin\left(\theta\right) \left[\dfrac{M_\text{rod} l}{2}+M_\text{sphere} \left(l+R\right)\right] \\ &=\left(9.8\right) \sin\left(30 \text{\textdegree}\right) \left[\dfrac{\left(1\right)\left(1\right)}{2} + \left(0.5\right) \left(1+0.30\right)\right] \\ &=5.64 \text{ N}\cdot \text{m} \end{align*}

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