An object with a mass of m = 5.3 kg is attached to the free end of a light strin

Question

An object with a mass of m = 5.3 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.255 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.
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

here,

mass of block, mb = 5.3kg
radius of wheel, R =0.255 m
mass of wheel, mw = 3 kg

Suspended length , H = 6.80m

Moment of inertia of wheel, I
I = 0.5 * M *R^2
I = 0.5 * 3 *(0.255)^2
I = 0.098 kg.m^2

Part A:
From newton second law of motion,
Fnet = mass * acceleration = ma
T - mg = ma (T is Tension in stinrg)
T = m(g+a) ----------------(1)

also,
Torque due rotation of wheel,
t = Tension in string * Perpendicular distance
t = T * R
t = (mg + ma)*R

also, Torque, t = moment of inertia * angular acceleration
(mg + ma)*R = I * Alpha
(mg + ma)*R = I * a/R (lpha = a/R)

a = mg/(I/R^2 + m) ----------------------(2)
a = (5 * 9.8) / ((0.098/0.255^2) + 5)
a = 7.53 m/s^2
the magnitude of the acceleration of the object is 7.53 m/s^2

Tension in string from eqn 1 :
T = mg + ma
T = 5(9.8 + 7.53)
T = 86.65 N

the tension in the string is 86.65 N.

From Third eqn of motion, Solving for Velocity, v
v^2 = 2aH
v = sqrt(2*a*H)
v = sqrt(2 * 7.53 * 6.80)
v = 10.12 m/s

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