A small ball of mass 1.0 kg is attached to one end of a 1.0 m long massless string and the other end of the string is hung from a point 0. When the resulting pendulum is making 30° from the vertical, what is the magnitude of net torque about the point
of suspension? [Take g = 10 m/s?]
Answers
Answered by
6
Answer:
Hey mate here is your answer....
We know that torque
τ
=
→
r
×
→
F
τ=r→×F→
from the fig we can see that torque along the suspension point will be
τ
=
→
r
×
→
T
+
→
r
×
→
F
τ=r→×T→+r→×F→
=
r
T
sin
0
+
r
M
g
sin
30
0
=rTsin0+rMgsin300
=
10
1
2
=
5
N
−
m
=1012=5N−m
Hop It's helpful for you ...
♥️♥️♥️ thanks
Answered by
4
Answer:
5 Nm
Explanation:
net torque T = force x perpendicular distance = mg.lsin∝ = 10sin30
T = 5 Nm
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