A uniform disk of mass M = 4.5 kg and radius R = 0.52 is mounted on a motionless

Question

A uniform disk of mass M = 4.5 kg and radius R = 0.52 is mounted on a motionless, fixed horizontal axis. A box with mass m = 0.80 kg hangs form a massless cord that is wrapped around the rim of the disk and is released from area. a. What is the direction of the direction of the angular velocity vector of the disk? b. Find the acceleration of the falling block. c. Find the tension in the cord. d. Find the angular acceleration of the disk. e. If released from rest, what is the speed of the block after it falls 1.5 m?

Explanation / Answer

A) the direction of the angular velocity vector of a disk is determined by using right hand rule. And the direction of angular velocity vector is outward from the disk.

B) acceleration of a block is nothing but the acceleration due due to gravity = 9.8 m/s2

C) tension in the cord = m x g

= 0.80 x 9.8

= 7.84 N

d) angular acceleration of a disk

T = r x F

= 0.52 x 7.84 = 4.07 N-m

T = I x a

Where,

I = moment of inertia (I for the disk = (1/2)mR2)

= ½ x 4.5 x 0.522

   = 0.608

a = angular acceleration

a = T/I

= 4.07 / 0.608

= 6.694 rad/s2 ans

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