Question
At t1 = 2.00 s, the acceleration of a particle in counter clockwise circular motion is
It moves at constant speed. At time t2 = 5.00 s, the particle's acceleration is 
What is the radius of the path taken by the particle if t2 - t1 is less than one period?
At t1 = 2.00 s, the acceleration of a particle in counter clockwise circular motion is
It moves at constant speed. At time t2 = 5.00 s, the particle's acceleration is What is the radius of the path taken by the particle if t2 - t1 is less than one period?
Answer: Option A
:
A
Att1=2s,→a=(5√3ms2)^i+5^j
Att2=5s,→a=(5ms2)^i−(5√3ms2)^i+^j
Given the body is moving in anti-clockwise direction
Since,the speed is constant, net acceleration will always be directed toward the centre and since it is
This will be the only path possible using this, we can calculate ω as we can find the change in angle
tanθ1=15√3=13
tanθ2=3√35=√3
∴θ1=30∘
⇒θ2=60∘
∴ We can see that angle between the vectors is 90∘ which means particle would have covered an angle of 360−90=270∘=3π/2radin3s
∴ω=ΔθΔt=3π23=π2rad/s
a=ω2r
r=aω2=√(5√3)2+52(π2)2=√100π24
=4×10π2=40π2
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:
A
Att1=2s,→a=(5√3ms2)^i+5^j
Att2=5s,→a=(5ms2)^i−(5√3ms2)^i+^j
Given the body is moving in anti-clockwise direction
Since,the speed is constant, net acceleration will always be directed toward the centre and since it is
This will be the only path possible using this, we can calculate ω as we can find the change in angle
tanθ1=15√3=13
tanθ2=3√35=√3
∴θ1=30∘
⇒θ2=60∘
∴ We can see that angle between the vectors is 90∘ which means particle would have covered an angle of 360−90=270∘=3π/2radin3s
∴ω=ΔθΔt=3π23=π2rad/s
a=ω2r
r=aω2=√(5√3)2+52(π2)2=√100π24
=4×10π2=40π2
Was this answer helpful ?
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