A nail is located at a certain distance vertically below the point of suspension of a simple pendulum of length 1m. The pendulum is released from the horizontal position. If it rotates in a vertical circle with nail as centre, the distiance of nail from point of suspension is
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54
the answer is-
conversion of P.E. of pendulum bob at A into K.E. AT L'
1/2mv²L=mgh=mg(L−SB)=mg(L−Lcos θ )
or
v²L=2gL(1−cos θ )
or
5gr=2gL(1−cos θ )
or
r=25×(1−cos60)=15=0.2
so,
y=L−r=1−0.2=0.8m
If it rotates in a vertical circle with nail as centre, the distiance of nail from point of suspension is 0.8 m
conversion of P.E. of pendulum bob at A into K.E. AT L'
1/2mv²L=mgh=mg(L−SB)=mg(L−Lcos θ )
or
v²L=2gL(1−cos θ )
or
5gr=2gL(1−cos θ )
or
r=25×(1−cos60)=15=0.2
so,
y=L−r=1−0.2=0.8m
If it rotates in a vertical circle with nail as centre, the distiance of nail from point of suspension is 0.8 m
Answered by
3
Answer:
0.8
Explanation:
the ANSWER IS 0.8
1/2MV^2=MGH=MG(L-SB)
V^2=2GL(1-COSTHETA)
Y=1-R=1-0.2=0.8
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