A thin, light string is wrapped around the rim of a 4.25 kg solid uniform disk t

Question

A thin, light string is wrapped around the rim of a 4.25 kg solid uniform disk that is 33.0 cm in diameter. A person pulls on the string with a constant force of 97.50 N tangent to the disk, as shown in the figure below (Figure 1) . The disk is not attached to anything and is free to move and turn.

A)Find the angular acceleration of the disk about its center of mass.

B)Find the linear acceleration of its center of mass.

C)If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the angular acceleration of the disk about its center of mass?

D)If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the linear acceleration of its center of mass

Explanation / Answer

a)

moment of inertia = I = (1/2)*M*R^2

torque = I*alfa

R*F = ( (1/2)*M*R^2 ) alfa

F = ( (1/2)*M*R ) *alfa

R =radius = 33/2 = 16.5 cm = 0.165 m

97.5 = ((0.5*4.25*0.165) )*alfa

alfa = 278.1 rad/s^2 <<-----------answer

b)

acceleration = a = R*alfa = 0.165*278.1 = 45.88 m/s^2

c)

moment of inertia = I = M*R^2

torque = I*alfa

R*F = ( M*R^2 ) alfa

F = (M*R) *alfa

R = radius = 33/2 = 16.5 cm = 0.165 m

97.5 = (4.25*0.165)*alfa

alfa = 139.04 rad/s^2 <<-----------answer

d)

acceleration = a = R*alfa = 0.165*139.04 = 22.94 m/s^2

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