A 0.420 kg pendulum bob passes throughthe lowest part of its path at a speed of
ID: 1723077 • Letter: A
Question
A 0.420 kg pendulum bob passes throughthe lowest part of its path at a speed of 3.30 m/s. (a) What is the tension in the pendulum cableat this point if the pendulum is 80.0 cm long?1 N
(b) When the pendulum reaches its highest point, what angle doesthe cable make with the vertical?
2°
(c) What is the tension in the pendulum cable when the pendulumreaches its highest point?
3 N (a) What is the tension in the pendulum cableat this point if the pendulum is 80.0 cm long?
1 N
(b) When the pendulum reaches its highest point, what angle doesthe cable make with the vertical?
2°
(c) What is the tension in the pendulum cable when the pendulumreaches its highest point?
3 N
Explanation / Answer
(a)
from the given problem we casn see that at thelowest point on the path the net upward force is
Fup = T - m g
= m (vt2 / r)
so the tension will be
T = m [g + (vt2/ r)]
= ......... N
(b)
according to the conservation of mechanicalenergy we get that
(KE + PEg)f = (KE +PEg)i
so as the bobgoes from the lowest to the highest pointwe get that
0 + m g [L (1 - cosmax)] = (1 /2) m vi2 + 0
cosmax = 1 -(vi2 / 2 g L)
max = ..........o
(c)
at the highest point on the path the bob is is atrest and the net radial force will be
Fr = T - m gcosmax
= m(vt2 / r)
= 0
T = m g cosmax
= ........ N
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