Question 590: Centre of Mass and Rotational Motion

Question

The moment of inertia of a circular disc of radius $$2 \mathrm{~m}$$ and mass $$1 \mathrm{~kg}$$ about an axis XY passing through its centre of mass and perpendicular to the plane of the disc is $$2 \mathrm{~kg} \mathrm{~m}^2$$. The moment of inertia about an axis parallel to the axis $$\mathrm{XY}$$ and passing through the edge of the disc is

Accepted Answer

$$\begin{aligned} & \mathrm{I}=1.2 \mathrm{~kg}-\mathrm{m}^2, \quad 	ext { K.E. }=1500 \mathrm{~J}, \alpha=25 \mathrm{rad} / \mathrm{s}^2 \ & 	ext { K.E. }=\frac{1}{2} \mathrm{I} \omega^2 \ & 	herefore 1500=\frac{1}{2} 	imes 1.2 	imes \omega^2 \ & 	herefore \omega^2=\frac{2 	imes 1500}{1.2}=2500 \quad 	herefore \omega=50 \mathrm{rad} / \mathrm{s} \ & \quad \omega=\omega_0+\alpha \mathrm{t} \ & 	herefore 50=0+25 \mathrm{t}=25 \mathrm{t} \quad 	herefore \mathrm{t}=2 \mathrm{~s}\end{aligned}$$

Answer

$$6 \mathrm{~kg} \mathrm{~m}^2$$

Answer

$$4 \mathrm{~kg} \mathrm{~m}^2$$

Answer

$$10 \mathrm{~kg} \mathrm{~m}^2$$

Answer

$$8 \mathrm{~kg} \mathrm{~m}^2$$

Centre of Mass and Rotational Motion

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