4.3 Solutions
177
(b) In coming down to the bottom of the hemisphere loss of potential
energy = mgh = mg R. Gain in kinetic energy = (7/10) mv 2 .
∴
7
10
mv
2
= mg R
or
mv 2
R
=
10mg
7
The normal force exerted by the small sphere at the bottom of the large
sphere will be
N = mg +
mv 2
R
= mg +
10mg
7
=
17mg
7
4.50 Work done W = τ θ = I αθ =
I ω 2
2
Along the diameter for hoop, I = m R 2 /2, while for the solid sphere, hollow
sphere and the disc, I = (2/5) m R 2 , (2/3) m R 2 and (1/4) m R 2 , respectively,
maximum work will have to be done to stop the hollow sphere, ω being identical as it has the maximum moment of inertia.
4.51 Work done W = τ θ = I αθ =
I ω 2
2
=
J 2
2I
where we have used the formula J = I ω. Maximum work will have to be
done for the disc since I is the least, τ being identical.
4.52 W =
1
2
I ω
2
=
1
2
(I ω)ω =
1
2
J ω
Since J and ω are the same for all the four objects, work done is the same.
4.53 τ = I α = I
a
R
=
Mg R sin θ
1 +
R 2 /k 2
For solid sphere, hollow sphere, solid cylinder and hollow cylinder the quantity 1 + (R 2 /k 2 ) is 7/2, 5/2, 3, 2, respectively. Therefore τ will be least for
solid sphere.