A pendulum of length 1m is released from theta =60 degree . The

1.	A pendulum of length 1m is released from theta =60 degree . The rate of change of speed of the bob at theta = 30 degree is
Answer» We know that Rate change of speed, dv/dt= tangential acceleration =tangential force/mass When θ=30^∘, tangential force is mgsinθ Therefore F=mgsin30 ma=mgsin30 a=mgsin30/m =10sin30 =10×1/2 =5m/s² So, The rate of change of speed of the bob at theta = 30 degree is 5m/s²

A pendulum of length 1m is released from theta =60 degree . The rate of change of speed of the bob at theta = 30 degree is

Answer»

We know that Rate change of speed, dv/dt= tangential acceleration =tangential force/mass

When θ=30^∘, tangential force is mgsinθ

Therefore F=mgsin30

ma=mgsin30

a=mgsin30/m

=10sin30

=10×1/2

=5m/s²

So, The rate of change of speed of the bob at theta = 30 degree is 5m/s²