126
3 Rotational Kinematics
Fig. 3.17
The tension in the string at C will be
T =
mv 2
L
+ mg cos θ
(3)
By problem, T = mg
(4)
Combining (2), (3) and (4) we get cos θ =
1
3
or θ = cos
−1
1
3
.
3.29 At the top of the sphere, v is in the horizontal direction and the frictional force
acts upwards. The condition that the motorcyclist may not fall is
Friction force = Weight
μ
mv 2
r
= mg
v =
gr
μ
=
9.8 × 10
0.8
= 11 m/s
3.30 At the lowest point A, Fig. 3.18, the tension in the string is
T A =
mv 2
A
L
+ mg
(1)
where v A is the velocity at point A.
Fig. 3.18
3 Rotational Kinematics
Fig. 3.17
The tension in the string at C will be
T =
mv 2
L
+ mg cos θ
(3)
By problem, T = mg
(4)
Combining (2), (3) and (4) we get cos θ =
1
3
or θ = cos
−1
1
3
.
3.29 At the top of the sphere, v is in the horizontal direction and the frictional force
acts upwards. The condition that the motorcyclist may not fall is
Friction force = Weight
μ
mv 2
r
= mg
v =
gr
μ
=
9.8 × 10
0.8
= 11 m/s
3.30 At the lowest point A, Fig. 3.18, the tension in the string is
T A =
mv 2
A
L
+ mg
(1)
where v A is the velocity at point A.
Fig. 3.18