a mass M= 1.50kg is suspended with a very light string which is tied to and wrap

Question

a mass M= 1.50kg is suspended with a very light string which is tied to and wrapped around the circumference of a uniform solid disk of mass 2M (exact) and radius R= .250 m. the disk is mounted on a very low friction bearing passing through its center of mass. the bottom of the suspended mass is 3.00 m above the floor of the lab. if the mass M is released from rest, how long with it take to strike the floor?
Given:
mass of the disk: 2M= 3.00kg
mass of the box: M= 1.50kg
Radius: R= .250m
height: h= 3.00m

Explanation / Answer

Given Mass of the box, M = 1.50 kg Mass of the disk, m = 3.0 kg Radius of the disk , R = 0.25 m Net force acting on the box is M a = M g - T     T = M g - Ma       ---1 Here T is the tension in the string Torque acting on the disk is = I R * T = I R(M g - Ma  ) = (1/2 mR^2 / 2) (a / R)        [From eq 1] M(g - a) = m a /2 1.50 kg (9.8 m/s^2 - a) = (3.0 kg *a )/ 2 a = 4.9 m/s^2    is the acceleration of the disk as it falls Box is relaesed from height , h = 3.0 m, then the time taken to reach ground is 3.0 m = vi t + 1/2 g t^2 3.0 m = 0 + 0.5 *(9.8 m/s^2) * t^2 t = 0.78 s Net force acting on the box is M a = M g - T     T = M g - Ma       ---1 Here T is the tension in the string Torque acting on the disk is = I R * T = I R(M g - Ma  ) = (1/2 mR^2 / 2) (a / R)        [From eq 1] M(g - a) = m a /2 1.50 kg (9.8 m/s^2 - a) = (3.0 kg *a )/ 2 a = 4.9 m/s^2    is the acceleration of the disk as it falls Box is relaesed from height , h = 3.0 m, then the time taken to reach ground is 3.0 m = vi t + 1/2 g t^2 3.0 m = 0 + 0.5 *(9.8 m/s^2) * t^2 t = 0.78 s

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