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Standard IX

Physics

Second Law of Motion

A thin rod of...

Question

A thin rod of length 1m is fixed in a vertical position inside a train, which is moving horizontally with constant acceleration $4 m s^{- 2}$ . A bead can slide on the rod, and friction coefficient between them is 1/2. If the bead is released from the rest at the top of the rod, find the time when it will reach at the bottom.

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Solution

If mass of bead is m
mg - µN=ma
Now, N = pseudo force due to acceleration of train, which is equal to 4m Newton.
=>   mg-µ(4m)=ma
=>  g-4µ=a
=>  a=8ms^-2
Using, second equation of motion:
1= 0.5(8) t²
T= 0.5sec

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If mass of bead is mmg - µN=maNow, N = pseudo force due to acceleration of train, which is equal to 4m Newton.=> mg-µ(4m)=ma=> g-4µ=a=> a=8ms-2Using, second equation of motion:1= 0.5(8) t2T= 0.5sec

If mass of bead is m
mg - µN=ma
Now, N = pseudo force due to acceleration of train, which is equal to 4m Newton.
=> mg-µ(4m)=ma
=> g-4µ=a
=> a=8ms^-2
Using, second equation of motion:
1= 0.5(8) t²
T= 0.5sec