A thin, light horizontal string is wrapped around the rim of a 4.00-kg solid uni

Question

A thin, light horizontal string is wrapped around the rim of a 4.00-kg solid uniform disk that is 30.0 cm in diameter. A box is connected to the right end of the string and moves to the right along the ground horizontally with no friction. The box is subjected to a horizontal force of magnitude F =100. 0 N parallel to the ground. The box has mass m = 1.00 kg. The disk is rotates clockwise about a fixed axis attached to a steel structure bolted to the ground: (a)What is a, the linear acceleration magnitude the box? (b) What is the tension T in the string? (c)If the disk is replaced with a hollow thin walled cylinder of the same mass, and diameter, what will be the acceleration in part (a)?

Explanation / Answer

for box

Fnet = m*a

F - T = m*a

T = F - ma...................(1)

for disk

net torque = I*alpha

T*R = I*alpha............(2)

angular acceleration = alpha = a/R

a = linear acceleration

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

from 1 & 2

(F - m*a)*R = (1/2)*M*R^2*a/R

(F - m*a) = (1/2)*M*a

(100-(1*a)) = (1/2)*4*a

acceleration = a = 33.33 m/s^2

b)

T = F- ma = 100-(1*33.33) = 66.67 N

c)

for hollow thin walled cylinder I = M*R^2

from 1 & 2

(F - m*a)*R = M*R^2*a/R

(F - m*a) = M*a

(100-(1*a)) = 4*a

a = 20 m/s^2 <<-------answer

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