Answer:
(A) 9.7 N
(B) 15.4 N
(C) I = 0.0045 kg m^{2}
Explanation:
mass of text book (M1) = 2.09 kg
mass of book (M2) = 2.99 kg
diameter of the pulley (d) = 0.12 m
radius (r) = 0.06 m
distance moved (s) = 1.30 m
time (t) = 0.75 s
acceleration due to gravity (g) = 9.8 m/s^[2}
(a) what is the tension in the part of the cord attached to the text book?
the text book is moving horizontally, so the tension in this case becomes
tension = mass x acceleration
we can get the acceleration from s = ut + 0.5 at^{2}
since the books are initially at rest u = 0
s = 0.5 at^{2}
1.3 = 0.5 x a x 0.75^{2}
a = 4.643 m/s^[2}
tension (T1) = 2.09 x 4.643 = 9.7 N
(b) what is the tension in the part of the cord attached to the book?
the book is hanging vertically, so the tension in this case becomes
tension = m x ( g - a )
(g-a) is the net acceleration of the first book
tension (T2) = 2.99 x (9.8 - 4.643) = 15.4 N
(c) What is the moment of inertia of the pulley?
if the books were to move in the direction of the book, it will cause the pulley to rotate clockwise and if they were to move in the direction of the text book on the table the pulley will rotate in an anticlockwise direction. Taking clockwise rotation of the pulley to be negative while anticlockwise to be positive, we can say ( T2 - T1 )r = I∝
where ∝ is the angular acceleration of the pulley relative to its radial
acceleration, ∝ = \frac{a}{r}
( T2 - T1 )r = I\frac{a}{r}
I = \frac{(T2 - T1)r^{2}}{a}
I = \frac{(15.5 - 9.7)0.060^{2}}{4.643}
I = 0.0045 kg m^{2}
we apply the equation
