An object with a mass of m = 5.35 kg is attached to the free end of a light stri

Question

An object with a mass of m = 5.35 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 5.70 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

Explanation / Answer

a)

torque on the reel is T = I*alpha

I is the moment of inertia = 0.5*M*R^2 = 0.5*3*0.235^2 = 0.0828 Kg-m^2

alpha is the angular accelaration = a/R

but Torque T = R*F

F is the tension in the string

R*F = 0.0828*(a/R)

R^2*F = 0.0828*a

0.235^2*F = 0.0828*a

a = 0.235^2*F/0.0828

a = 0.67*F...........(1)

writing equation of motion for descending mass m is

mg - F = ma

5.35*9.8 - F = 5.35*a

F = (5.35*9.8)-(5.35*a)

F = (5.35*9.8)-(5.35*0.67*F)

tension is F = 11.5 N

b) accelaration is a

a = 0.67*F = 0.67*11.5 = 7.705 m/s^2

c) using kinematic equations

distance travelled is S = 5.7 m

initial speed is Vo = 0 m/sec

final speed is V

using

v^2 - vo^2 =2*a*S

v^2 - 0^2 = 2*7.705*5.7

v = 9.4 m/sec

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