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A body moves on a horizontal circular ro...

A body moves on a horizontal circular road of radius `r`, with a tangential acceleration `a_(t)`. The coefficient of friction between the body and the road surface Is `mu`. It begins to slip when its speed is `v`.
(i) `v^(2)=murg`
(ii) `mug=(v^(4)/(r^92))+a_(t))`
(iii) `mu^(2)g^(2)=(v^(4)/(r^(2)+a_(t)^(2))`
(iv) The force of friction makes an angle `tan^(-1)(v^(2)//a_(t)r)` with the direction of motion at the point of slipping.

A

`v^(2)=murg`

B

`mug=v^(2)/r+a_(T)`

C

`mu^(2)g^(2)=v^(4)/r^(2)+a_(T)^(2)`

D

The force of friction makes an angle `tan^(-1)((v^(2))/(a_(T)xxr))` with direction of motion of point of slipping.

Text Solution

Verified by Experts

The correct Answer is:
C, D
At time of slipping `f=mumg`
`f cos theta=ma_(T), f^(2)=(ma_(T))+((mv^(2))/r)`
`(mumg)^(2)=(ma_(T))^(2)+((mv^(2))/r)^(2)`
`rArr mu^(2)g^(2)=a_(T)^(2)+((v^(4))/r^(2))`,Also `tan theta=v^(2)/(a_(T)r)`
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