A block of mass m = 2.0 kg is connected to a massless inextensible string with a

Question

A block of mass m = 2.0 kg is connected to a massless inextensible string with a negligible thickness. The string is attached to and wrapped many times around a uniform circular pulley of mass m (same as the block mass) and radius R = 0.45 m. The pulley can rotate around its axis freely. The system of the pulley, the block, and the string is released from rest at time 0. At later time, t, the block has descended by a height of H, and the speed of the block is 5.5 m/s. Assume that the string does no slip on the pulley.

Explanation / Answer

a)

Here , let the height is H

Now, using conservation of energy

kinetic energy of pulley + kinetic energy of block = decrease in potential energy of block

0.5 * I * w^2 + 0.5 * m * v^2 = m*g*H

0.5 *( 0.5 *m *R^2 *(5.5/R)^2 + m * 5.5^2) = m * 9.8 * H

solving for H

H = 2.32 m

the height H is 2.32 m

b)

as v = R*w

w = 5.5/0.45

w = 12.22 rad/s

angular velocity is 12.22 rad/s anticlockwise direction

c)

let the tension in the string is T

using work energy theorum

(mg - T) * H = 0.5 * m * v^2

(2 * 9.8 - T) * 2.32 = 0.5 * 2 * 5.5^2

solving for T

T = 6.56 N

the tension in the string is 6.56 N

d)

angular momentum of block = m*v*R

angular momentum of block = 2 * 0.45 * 5.5

angular momentum of block = 4.95 Kg.m^2/s

the angular momentum of block is 4.95 Kg.m^2/s

as L = r X p

it is OUT OF THE PAGE

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