A 0.300 kg pendulum bobpasses through the lowest part of its path at a speedof 3
ID: 1742987 • Letter: A
Question
A 0.300 kg pendulum bobpasses through the lowest part of its path at a speedof 3.30 m/s. (a) What is the tension in the pendulum cable at this point if thependulum is 80.0 cm long?N
(b) When the pendulum reaches its highest point, what angle doesthe cable make with the vertical?
°
(c) What is the tension in the pendulum cable when the pendulumreaches its highest point?
N (a) What is the tension in the pendulum cable at this point if thependulum is 80.0 cm long?
N
(b) When the pendulum reaches its highest point, what angle doesthe cable make with the vertical?
°
(c) What is the tension in the pendulum cable when the pendulumreaches its highest point?
N
Explanation / Answer
(a) Two forces act on the pendulum bob:(i) its weight, mg, downwards and
(ii) tension, T, in the cable, upwards.
The resultant of these two forces in the upwards direction,
T - mg = mv^2/r, where v = velocity of the bob and r = cablelength.
T = m(g + v^2/r)
= (0.300)[9.8 + (3.30)^2/0.800] ----> 80.0 cm =0.800 m
= 7.024 N
(b) Let h = vertical distance of the highest point from the lowestpoint, and
= angle made by the cable with the vertical at thatinstant.
Then, h = r - r cos = r(1 - cos) =2rsin^2(/2)
h = (2r)*sin^2(/2) -----> ( 1 )
When the bob reaches the highest point, its K.E. at the lowestpoint gets converted to P.E. at the highest point.
So, (1/2)mv^2 = mgh
h = v^2/2g ---> ( 2 )
From eqns. ( 1 ) and ( 2 ),
(2r)*sin^2(/2) = v^2/2g
sin(/2) = v / 2(rg)
sin(/2) = 3.3 / 2(0.8)(9.8)
sin(/2) = 0.59
= 72.3°
(c) At the highest point, its velocity is zero. So no centripetalforce on it. Hence, T should equal the component of mg in itsdirection.
T = mg cos 72.3° = (0.3)(9.8) (0.030)
T = 0.088N -- -- Hope this helps.
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