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 - 0.275 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 6.80 m above the floor. (a) Determine the tension in the string (in N). (b) Determine the magnitude of the acceleration of the object (in m/s2). m/s2 (e) Determine the speed with which the object hits the floor (in m/s). m/s

Answer 1

Moment of Inertia of Solid disk;

J = (1/2) M r² = 0.5 * 3 * 0.275² = 0.1134 kg-m²

We will denote tension in string as "T" . Torque on disk will be T*r and

T *r = J *α ; where α is angular acceleration.

Also; we can write Tension

T = mg - ma ; where a is linear acceleration and a = α*r

(mg-ma)*r = J* α = J*a/r

(mg-ma)*r² = J*a

a(mr² +J) = mgr²

a =  mgr² / (mr² +J)

a = 4.8 * 9.81* 0.275² /(4.8*0.275² + 0.1134)

a = 7.47 m/s² (Answer for part b)

T = mg -ma = 4.8 *9.81 - 4.8 * 7.47 = 11.21 N (Answer for Part a)

As mass m will start falling with an acceleration of a = 7.47 m/s²

We will use v² = u² +2as ; in our case, u = 0, s= 6.8 m and a = 7.47

v² = 0² + 2*7.47 *6.8 = 101.59

v = sqrt(101.59) = 10.08 m/s (Answer for part c)