A uniform cylindrical turntable of radius 1.50 m and mass 31.1 kg rotates counte

Question

A uniform cylindrical turntable of radius 1.50 m and mass 31.1 kg rotates counterclockwise in a horizontal plane with an initial angular speed of 4 rad/s. The fixed turntable bearing is frictionless. A lump of clay of mass 2.27 kg and negligible size is dropped onto the turntable from a small distance above it and immediately sticks to the turntable at a point 1.40 m to the east of the axis.

(a) Find the final angular speed of the clay and turntable.

(b) Is mechanical energy of the turntable-clay system conserved in this process?
---Select--- no yes  

What, if any, is the change in internal energy?
J

(c) Is momentum of the system conserved in this process?
---Select--- No yes

What, if any, is the amount of impulse imparted by the bearing?

kg · m/s

Explanation / Answer

a) Moment of Inertia turnable, I = (1/2)*M*R^2

= (1/2)*31.1*1.5^2

= 34.9875 kg.m^2

w1 = 4*pi rad/s

Apply conservation of angular momentum

I2*w2 = I1*w1

(I + m*r^2)*w2 = I*w1

(34.9875 + 2.27*1.4^2)*w2 = 34.9875*4*pi

w2 = 34.9875*4*pi/(34.9875 + 2.27*1.4^2)

= 11.15 rad/s (counter clcokswise)

b) No.

KEi = 0.5*I*w1^2

= 0.5*34.9875*(4*pi)^2

= 2762.5 J

KEf = 0.5*I2*w2^2

= 0.5*(34.9875 + 2.27*1.4^2)*11.15^2

= 2541.4 J

change in internal energy = KEi - KEf

= 2762.5 - 2541.4

= 221.1 J

c) yes.

d) Impulse = change in momentum of the clay

= m*(v2 - v1)

= m*v2

= m*r*w2

= 2.27*1.4*11.15

= 35.4 kg.m/s

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