4.3 Solutions
179
Therefore (1) becomes
v = v 0 −
ω R
2
= v 0 −
v
2
∴ v =
2
3
v 0
(2)
Using (2) in (1)
2
3
v 0 = v 0 − μ gt
or t =
v 0
3 μg
(b) Work done W = K =
1
2
mv 2 −
1
2
mv 2
0
=
1
2
m
4
9
v
2
0 − v
2
0
= −
5
18
mv
2
0
4.57 Equation of motion is
ma = mg − 2T
(1)
where ‘a’ is the linear acceleration and T the tension in each thread.
Torque I α = 2T r
(∵ there are two threads)
1
2
mr
2
α = 2T r
or α =
4T
mr
(2)
As both the cylinders are rotating,
a = 2αr =
8T
m
(3)
or ma = 8T
(4)
Using (4) in (1) we get
T =
1
10
mg