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The simple 2Kg pendulum is released from...

The simple `2Kg` pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin aat `B` and continues in the smaller arc in the vertical plane. Calculate the magnitude of the force `R` supported by the pin at `B` when the pendulum passes the position `theta=30^(@),(g=9.8m//s^(2))`

Text Solution

Verified by Experts

The correct Answer is:
D
Speed of bob in the given position,
`v=sqrt(2gh)`
Here, `h=(400+400cos30^(@))mm`
`=746mm=0.746m`
:. `v=sqrt(2xx9.8xx0.746)=3.82m//s`

Now `T-mgcostheta=(mv^(2))/(r)`
`T=2xx9.8xxcos30^(@)+(2xx(3.82)^(2))/((0.4))`
or `T=90N`
:. `R=Tsin30^(@)=45N`
`T'=Tcos30^(@)`
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