8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

×

an angle θ = 35◦ beneath the horizontal. What is the angular acceleration of the

rod immediately after it is released?

pivot

x




l/2

rod

l (^) m g
Answer: The moment of inertia of a rod of mass m and length l about an axis,
perpendicular to its length, which passes through one of its ends is I = (1/3) m l^2
(see question 8.3). Hence,
5.3 1.3^2
I =
3
= 2.986 kg m^2.
The angular equation of motion of the rod is
I α = τ,
where α is the rod’s angular acceleration, and τ is the net torque exerted on the
rod. Now, the only force acting on the rod (whose line of action does not pass
through the pivot) is the rod’s weight, m g. This force acts at the centre of mass
of the rod, which is situated at the rod’s midpoint. The perpendicular distance x
between the line of action of the weight and the pivot point is simply
x =
l
cos θ =
1.3 × cos 35 ◦
2 2
= 0.532 m.
Thus, the torque acting on the rod is
τ = m g x.
It follows that the rod’s angular acceleration is written
τ m g x
α = =
I I

5.3 × 9.81 × 0.532
2.986
= 9.26 rad./s^2.