A 3.0 kg, 25-cm-diameter turntable rotates at 33 1/3 rpm on frictionless bearing

Question

A 3.0 kg, 25-cm-diameter turntable rotates at 33 1/3 rpm on frictionless bearings. A thin, 1.0 kg, 25 cm-long block falls from above, hits the turntable lying right across the middle and sticks. a) What is the turntable's angular velocity, in rpm, just after this event? b) A second 12.5 cm-long, 0.50 kg block falls and sticks with one end on the center of an identical turntable rotating at 33 1/4 rpm. The other end of the block is at the edge of the turntable. What is linear speed of the outer end of the block? c) Explain in words how you solved this problem. You can complete this by annotating your work in parts (a) and (b).

Explanation / Answer

Here the important point to remember is that.. the angular momentum should be conserved.

A) the initial angular momentum should be equal to the final angular momentum. Therefore, after the mass falls and sticks the new angular velocity would be

=m1v1/m2

= 3x33.3/4 rpm

= 25 rpm

B) similarly as above in this case the angular momentum will be

= 3x33.3/3.5 rpm

= 28.57 rpm

Now the block alligns itself along the radius and hence the linear velocity at the outer edge will be

V= rw

= 3.57 m/sec

C) The above two problems can be solved by conserving the angular momentum.

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