An object with a mass of m = 4.80 kg is attached to the free end of a light stri

Question

An object with a mass of m = 4.80 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.235 m and mass of M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest 5.30 m above the floor.

(a) Determine the tension in the string.
N

(b) Determine the magnitude of the acceleration of the object.
m/s2

(c) Determine the speed with which the object hits the floor.
m/s

(d) Verify your answer to part (c) by using the isolated system (energy) model. (Do this on paper. Your instructor may ask you to turn in this work.)

Explanation / Answer

on mass m ,

m g - T = m a .... (i)

on disk , torque = I alpha

R T = (M R^2 / 2) (a/R)

T = M a / 2 .... (ii)

from (i) and (ii),

4.80 x 9.81 = (4.80 + 1.50) a

a = 7.47 m/s^2

(A) T = (3 x 7.47)/2 = 11.2 N

(B) a = 7.47 m/s^2

(C) Applying vf^2 - vi^2 = 2 a d

v^2 - 0 = 2 (7.47)(5.30)

v = 8.90 m/s

(d) Applying energy conservation,

m g h = m v^2 /2 + I w^2 / 2

m g h = m v^2 /2 + (M R^2 /2)(v/R)^2 /2

m g h = m v^2 / 2 + M v^2 / 4

4.80 x 9.81 x 5.30 = v^2 (4.80/2 + 3/4)

v =8.90 m/s

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