10. In the figure below, two 5.60 kg blocks are connected by a massless string o

Question

10. In the figure below, two 5.60 kg blocks are connected by a massless string over a pulley of radius 1.95 cm and rotational inertia 6.60 10-4 kg m2. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley's axis is frictionless. When this system is released from rest, the pulley turns through 1.10 rad in 80.0 ms, and the acceleration of the blocks is constant. (a) What is the magnitude of the pulley's angular acceleration? rad/s2 (b) What is the magnitude of either block's acceleration? m/s2 (c) What is the string tension T1? N (d) What is the string tension T2? N

11.The thin uniform rod in the figure below has length 2.5 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle = 40° above the horizontal. Use the principle of conservation of energy to determine the angular speed of the rod as it passes through the horizontal position. rad/s

Explanation / Answer

Here ,

mass , m = 5.6 Kg

I = 6.6 *10^-4 Kg.m^2

radius ,r = 1.95 cm = 0.0195 m

theta = 1 rad

t = 80 ms

a)

let the angular acceleration is alpha

theta = 0.5 * alpha * t^2

1 = 0.5 * alpha * 0.080^2

alpha = 312.5 rad/s^2

magnitude of the pulley's angular acceleration is 312.5 rad/s^2

b)

magnitude of block's acceleration = r * alpha

magnitude of block's acceleration = 312.5 * 0.0195 m/s^2

magnitude of block's acceleration = 6.1 m/s^2

c)

Now , for the tension T1

T1 = m1 * a

T1 = 5.6 *6.1 N

T1 = 34.16 N

d)

for T2

T2 = 5.6 * (9.8 - 6.1)

T2 = 20.7 N

the tension T2 is 20.7 N

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