4.3 Solutions
175
mg − N = m
d 2
dt 2 (a − a cos θ)
or mg − N = ma
cos θ
dθ
dt
2
+ sin θ
d 2 θ
dt 2
(1)
The work–energy theorem gives
mga(1 − cos θ) =
1
2
m
dθ
dt
2
a 2
3
+ a
2 sin
2
θ
where the square bracket has been written using the parallel axis theorem.
dθ
dt
2
=
6g(1 − cos θ)
a(1 + 3 sin
2
θ)
(2)
∴
d 2 θ
dt 2 =
3g
a
sin θ(7 − 6 cos θ − 3 sin
2
θ)
(1 + 3 sin
2
θ) 2
(3)
By substituting
dθ
dt
2
and
d 2 θ
dt 2
from (2) and (3) in (1), the reaction N is
obtained as a function of θ . When the rod is about to strike the floor,
θ =
π
2
;
dθ
dt
2
=
3g
2a
and
d 2 θ
dt 2 =
3g
4a
Thus the reaction from (1) will be
N = m
g −
3g
4
or
1
4
mg
4.47 (a) For α net = 0, the two torques which act in the opposite sense must be
equal (Fig. 4.30), i.e.
τ 1 = τ 2
or m 1 g R 1 = m 2 g R 2
m 2 =
m 1 R 1
R 2
=
25 × 1.2
0.5
= 60 kg
(b) (i) a 1 = α R 1 , a 2 = α R 2
(1)
as R 1 > R 2 , a 1 > a 2
(ii) Equations of motion are