A block of mass m = 1.14 kg is suspended above the ground at a height h = 16.0 m

Question

A block of mass m = 1.14 kg is suspended above the ground at a height h = 16.0 m by a spool with two arms. The spool-with- arms arrangement is a combination of a solid uniform cylinder of mass M = 3.79 kg and radius R = 0.260 m, and two rods, each of length l = 0.296 m and mass mrod = 0.310 kg. The system is released from rest.

a) How fast will the block be moving just before it hits the ground? Answer in units of m/s.
b) What was in m/s2 the acceleration of the mass m then?
c) What was then, in N . m, the net torque on the cylinder?
d) What was in N the magnitude of the tension on the string?

Explanation / Answer

given,

mass of the block = 1.14 kg

height above the ground of the block = 16 m

mass of the cylinder = 3.79 kg

radius of the cyliner = 0.26 m

mass of rod = 0.31 kg

radius lenth of the rod = 0.296 m

moment of inertia of the cyliner = 0.5 * M * R^2

moment of inertia of the cyliner = 0.5 * 3.79 * 0.26^2

moment of inertia of the cyliner = 0.128 kg m^2

moment of inertia of the rod = (1/12) * M * L^2

moment of inertia of the rod = (1/12) * 0.31 * 0.296^2

moment of inertia of the rod = 0.00226 kg m^2

total moment of inertia = moment of inertia of the cyliner + 2 * moment of inertia of the rod

total moment of inertia = 0.128 + 2 * 0.00226

total moment of inertia = 0.13252 kg m^2

torque is due to weight of the block = weight of the block * radius of the cylinder

torque is due to weight of the block = 1.14 * g * 0.26

torque is due to weight of the block = 2.9 Nm

he net torque on the cylinder = 2.9 Nm

acceleration of the rod + cylinder system = torque / moment of inertia

acceleration of the rod + cylinder system = 2.9 / 0.13252

acceleration of the rod + cylinder system = 21.883 rad/sec^2

linear acceleration = 21.883 * 0.26

acceleration of the mass m then = 5.69 m/s^2

by third equation of motion

final angular velocity^2 = initial angular velocity^2 + 2 * angular acceleration * s

final angular velocity^2 = 0 + 2 * 21.883 * (16 / 0.26)

final angular velocity = 51.897 rad/sec

linear velocity = 51.897 * 0.26

block be moving just before it hits the ground with velocity = 13.493 m/s

net force on the block = mg - T

so,

mg - T = ma

1.14 * 9.8 - T = 1.14 * 5.69

magnitude of the tension on the string = 4.6854 N

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