An object with a mass of m = 4.6 kg is attached to the free end of a light strin

Question

An object with a mass of m = 4.6 kg is attached to the free end of a light string wrapped around a reel of radiusR = 0.240 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 5.90 m above the floor.

a) determine the tension in the string.

b) determine the magnitudde of the acceleration of the object.

c) determine the speed with which the object hits the floor.

Explanation / Answer

let F be the tension in the string

then net force Fnet = m*a = mg-F

F = mg-ma

torque T = I*alpha

I is the moment of inertia = 0.5*M*R^2

and alpha is the angular accelaration = a/R

T = R*F = 0.5*M*R^2*(a/R)

R^2*F = 0.5*M*R^2*a

F = 0.5*M*a

mg-ma = 0.5*M*a

mg = (0.5*M + m)*a

accelaration a = mg/(0.5M + m)

a = (4.6*9.81)/(0.5*3 + 4.6) = 7.4 m/s^2

then tension F = mg-ma = 4.6*(9.81-7.4) = 11.086 N

then A) F = 11.086 N

B) accelaration a = 7.4 m/s^2

C) initial speed is u = 0 m/s

acclearation a = 7.4 m/s^2

distance travelled is S = 5.9 m

then use v^2-u^2 = 2*a*S

v = Sqrt(2*a*S) = Sqrt(2*7.4*5.9) =9.4 m/s

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