A 2.10 kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter
is 0.170 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.900 s.(a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?
1 answer:
Answer:
Explanation:
Answer:
Explanation:
mass of book, m = 2.10 kg
diameter of pulley, = 0.170 m
radius of pulley, R = 0.085 m
mass of hanging book, m' = 3 kg
initial velocity, u = 0 m/s
distance, s = 1.2 m
time, t = 0.9 s
Let a be the acceleration of the system and T and T' is the tension in the string which is horizontal and vertical respectively.
Use second equation of motion
s = ut + 0.5 at²
1.2 = 0 + 0.5 x a x 0.9 x 0.9
a = 2.96 m/s²
(a) Use second equation of motion
T = ma
T = 2.10 x 2.96
T = 6.216 N
m'g - T' = m'a
3 x 9.8 - T' = 3 x 2.96
T' = 20.52 N
(b) Let the moment of inertia of the pulley is I.
So, (T' - T)R = I x α
(20.52 - 6.216) x 0.085 = I x 2.96 / 0.085
I = 0.035 Kgm²
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