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Question: A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is \(2 \m...

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 2 m s−22 \mathrm {~m} \mathrm {~s} ^ { - 2 } . The magnitude of net acceleration at this instant is :

A

5 m s−25 \mathrm {~m} \mathrm {~s} ^ { - 2 }

B

3.2 m s−23.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }

C

2 m s−22 \mathrm {~m} \mathrm {~s} ^ { - 2 }

D

4.3 m s−24.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }

Answer

2 m s−22 \mathrm {~m} \mathrm {~s} ^ { - 2 }

Explanation

Solution

Here r=10m,v=5 ms−1,a=2ms−2r = 10 m , v = 5 \mathrm {~ms} ^ { - 1 } , a = 2 m s ^ { - 2 }

ar=v2r=5×510=2.5 ms−2a _ { r } = \frac { v ^ { 2 } } { r } = \frac { 5 \times 5 } { 10 } = 2.5 \mathrm {~ms} ^ { - 2 }

The net acceleration is

a=ar2+at2=(2.5)2+22a = \sqrt { a _ { r } ^ { 2 } + a _ { t } ^ { 2 } } = \sqrt { ( 2.5 ) ^ { 2 } + 2 ^ { 2 } }

=10.25=3.2 ms−1= \sqrt { 10.25 } = 3.2 \mathrm {~ms} ^ { - 1 }