A 2.3 kg, 20-cm-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 diagonal, and stick. What is the turntable's angular velocity, inrpm, just after this event?

Question

Giovani Grimes

Physics

11 April, 02:36

A 2.3 kg, 20-cm-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 diagonal, and stick. What is the turntable's angular velocity, inrpm, just after this event?

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Answers (1)

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Israel Fowler · Answer 1

Answer

ω2=82.1 rpm

Explanation:

given required

m1=2.3 kg ω2=?

m2+m3=0.5kg

r1=10 cm

ω1=100 rpm

solution

Using the application of conservation of angular momentum we can solve as follows/

L before collision = L after collision

m1r²ω1 = (m1+m2+m3) r²ω2

2.3 kg*0.1² m²*100 rpm = (2.3 kg+0.5 kg) * 0.1²m²*ω2

2.3 rpm=0.028*ω2

ω2=82.1 rpm