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
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
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_a8718af38593777fe50149487e66032f3c6814b020160609-1956-jq2vma.png)
Since,the speed is constant, net acceleration will always be directed toward the centre and since it is
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_b229d2a20569860e367e32eae404ccb91fc7d76320160609-1956-1nhmv7f.png)
This will be the only path possible using this, we can calculate ω as we can find the change in angle
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_91e45806fcc6f8a305461535ed7f5800fe95062b20160609-1956-1vkz8vw.png)
tanθ1=15√3=13
tanθ2=3√35=√3
∴θ1=30∘
⇒θ2=60∘
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_359d0651e26092a2a8bf5b2b28b67b52444bff4f20160609-1956-13l6813.png)
∴ 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
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_a8718af38593777fe50149487e66032f3c6814b020160609-1956-jq2vma.png)
Since,the speed is constant, net acceleration will always be directed toward the centre and since it is
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_b229d2a20569860e367e32eae404ccb91fc7d76320160609-1956-1nhmv7f.png)
This will be the only path possible using this, we can calculate ω as we can find the change in angle
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_91e45806fcc6f8a305461535ed7f5800fe95062b20160609-1956-1vkz8vw.png)
tanθ1=15√3=13
tanθ2=3√35=√3
∴θ1=30∘
⇒θ2=60∘
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_359d0651e26092a2a8bf5b2b28b67b52444bff4f20160609-1956-13l6813.png)
∴ 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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