Question

A turntable has a radius R and mass M (considered as a disk) and is rotating at an angular velocity W0 about a frictionless vertical axis. A piece of clay is tossed onto the turntable and sticks d from the rotational axis. The clay hits with horizontal velocity component vc at right angle to the turntable’s radius, and in a direction that opposes the rotation. After the clay lands, the turntable has slowed to angular velocity W1. A) Find the mass of the clay mc. B) How much energy is lost during the collision process (i.e., due to dissipation and deformation of the clay)?

Answer #1

A)

I_{t} = Moment of inertia of turntable = (0.5) M
R^{2}

W_{o} = initial angular velocity of turntable

I_{c} = moment of inertia of clay = m_{c}
d^{2}

W_{1} = final angular velocity

using conservation of angular momentum

I_{t} W_{o} - m_{c} v_{c} d =
(I_{t} + I_{c}) W_{1}

I_{t} W_{o} - m_{c} v_{c} d =
(I_{t} + m_{c} d^{2}) W_{1}

I_{t} (W_{o} + W_{1}) = m_{c}
v_{c} d + m_{c} d^{2} W_{1}

(0.5) M R^{2} (W_{o} + W_{1}) =
m_{c} v_{c} d + m_{c} d^{2}
W_{1}

m_{c} = (0.5) M R^{2} (W_{o} +
W_{1}) /(v_{c} d + d^{2} W_{1})

b)

Energy lost is given as

E
= (0.5) (I_{t} W^{2}_{o} + m_{c}
v^{2}_{c} - (I_{t} + I_{c})
W^{2}_{1})

E
= (0.5) ((0.5) M R^{2} W^{2}_{o} +
m_{c} v^{2}_{c} - ((0.5) M R^{2} +
m_{c}d^{2}) W^{2}_{1})

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