70
2 Particle Dynamics
2.12 (a) The torque due to the external gravitational force on M 1 will be M 1 gr,
and the torque due to the external gravitational force on M 2 will be the
component of M 2 g along the string times r , i.e. (M 2 g sin θ)gr. Now,
these two torques act in opposite directions. Taking the counterclockwise
rotation of the pulley as positive and assuming that the mass M 1 is falling
down, the net torque is
τ = M 1 gr − (M 2 g sin θ)r = (M 1 − M 2 sin θ)gr
(1)
and pointing out of the page.
(b) When the string is moving with speed υ, the pulley will be rotating with
angular velocity ω = v/r , so that its angular momentum is
L pulley = I ω =
I v
r
and that of the two blocks will be
L M 1 = r M 1 v
L M 2 = r M 2 v
All the angular momenta point in the same direction, positive if M 1 is
assumed to fall. The total angular momentum is then given by
L total = v
(M 1 + M 2 )r +
I
r
(2)
(c) Using (1) and (2)
τ =
dL
dt
=
dυ
dt
(M 1 + M 2 )r +
1
r
= [M 1 − M 2 sin θ ] gr
The acceleration a =
dv
dt
=
[M 1 − M 2 sin θ ] g
(M 1 + M 2 ) +
1
r 2
2.13 (i) ma = F − mg sin θ (equation of motion, up the incline)
F = ma + mg sin θ = m(a + g sin θ)
= (1.0)(1 + 9.8 × 0.5) = 5.9 N (θ = 30
◦
)
(ii) ma = F + mg sin θ (equation of motion, down the incline)
∴ F = ma − mg sin θ = m(a − g sin θ)
= (1.0)(1 − 9.8 × 0.5) = −3.9 N
The negative sign implies that the force F is to be applied up the incline.
2.14 The displacement on the edge is measured by s while that on the floor by x. As
the mass m goes down the wedge the wedge itself would start moving towards