Radial & Tangential Acceleration for Non Uniform Circular Motion
A particle is...
Question
A particle is moving in a circle of radius r=2m. Its speed with time varies according to the relation v(t)=4−6t2. What is the magnitude of the total acceleration of the particle at t=2sec ? Use the value of √769=27.73 for calculation.
A
2.11m/s2
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B
3.46m/s2
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C
3.1m/s2
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D
1.5m/s2
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Solution
The correct option is B3.46m/s2 Given the variation of the speed of the particle is: v(t)=4−6t2
Diffrentiating both sides w.r.t time, we get: dvdt=0+12t3
Now, tangential acceleration at is the rate of change of speed with time,
i.e at=dvdt=12t3
At t=2s, at=128=32m/s2
Now speed at t=2s: v(t=2)=4−64=52m/s
So, centripetal acceleration at t=2sec ∴ac=v2r=(254)2=258m/s2
Total acceleration aT of particle is aT=√a2c+a2t=√62564+94=√7698 =27.738=3.46m/s2