A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s^2 in a circular path. If it slips when its total acceleration becomes 1 m/s. Find the angle through which it would have turned before it starts to slip.

Question

HARSHDEEP SINGH CHHABRA · Answer

When net acceleration=1 Tangential acceleration=0.6 and centripetal =0.8 Now w^2R=0.8 So w^2=0.8/r Tagential acceleration=0.6=rα So α=0.6/r Now as acceleration is constant, we can use w^2=2α theta You can get theta from here which equals 2/3rad

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A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s^2 in a circular path. If it slips when its total acceleration becomes 1 m/s. Find the angle through which it would have turned before it starts to slip.

A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s^2 in a circular path. If it slips when its total acceleration becomes 1 m/s. Find the angle through which it would have turned before it starts to slip.

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A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s^2 in a circular path. If it slips when its total acceleration becomes 1 m/s. Find the angle through which it would have turned before it starts to slip.

A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s^2 in a circular path. If it slips when its total acceleration becomes 1 m/s. Find the angle through which it would have turned before it starts to slip.

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