Lorraine Harvey

2022-01-06

A 2.0 kg, 20cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

abonirali59

The mas of turntable $M=2Kg$
Diameter of the turntable $d=20cm=0.2m$
Angular velocity $\omega =10r±=100×\frac{2\pi }{60}=10.47ra\frac{d}{s}$
Mass of the two blocks $m=500g=0.5kg$
To determine the turntable's angular velocity once the two blocks stick with it. When the two blocks collide, the moment of inertia of the entire system changes, causing the angular velocity of the system to change. The angular velocity of the turntable can be calculated using conservation of angular momentum:
${L}_{\in itial}={L}_{f\in al}$
${I}_{turntab\le }\omega ={I}_{block}{\omega }^{\prime }+{I}_{turntable}{\omega }^{\prime }+{I}_{block2}{\omega }^{\prime }$

${\omega }^{\prime }=\frac{{I}_{turntable}\omega }{{I}_{block}{\omega }^{\prime }+{I}_{turntable}{\omega }^{\prime }+{I}_{block2}{\omega }^{\prime }}$
${I}_{turntable}=M\frac{{r}^{2}}{2}=2×\frac{{0.1}^{2}}{2}=0.01$
${I}_{block1}=m{r}^{2}0.5×{0.1}^{2}=0.005={I}_{block2}$
${\omega }^{\prime }=\frac{0.01×10.47}{0.005+0.01+0.005}$
$=5.235ra\frac{d}{s}$
$=5.235×\frac{60}{2\pi }=50r±$

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