KVPY 2016 SA Question 9

Classical Mechanics Level 3

A car goes around a uniform circular track with radius R R at uniform speed v such that at every T T seconds it completes rotation. The magnitude of centrepetal accelaration is a c a_c . Now if the car goes uniformly around a larger circular track with acceleration 8 a c 8a_c and radius 2 R 2R . Then the time period is

T ′ T' = a T b \dfrac{aT}{b} where a a and b b are coprime positive integers. Find a + b a+b .

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The answer is 3.

Md Zuhair by Md Zuhair , 20, India

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2 solutions

Tapas Mazumdar
Feb 26, 2017

The time period T T is given by

T = 2 π ω T = \dfrac{2\pi}{\omega}

where ω \omega (angular velocity) = v R \ = \dfrac vR .

Now, we have

T = 2 π ω T ′ = 2 π ω ′ T = \dfrac{2\pi}{\omega} \\ T' = \dfrac{2\pi}{\omega '}

Which gives

T T ′ = ω ′ ω \dfrac{T}{T'} = \dfrac{\omega '}{\omega}

The centripetal acceleration a c a_c is given by

a c = R ω 2 a_c = R \omega^2

Now according to our problem

8 a c = 2 R ω ′ 2 8 a_c = 2R {\omega '}^2

Which gives

ω ′ 2 ω 2 = 4 ⟹ ω ′ ω = 2 \begin{aligned} & \dfrac{{\omega '}^2}{\omega^2} = 4 \\ \implies & \dfrac{\omega '}{\omega} = 2 \end{aligned}

So

T T ′ = 2 ⟹ T ′ = T 2 \dfrac{T}{T'} = 2 \implies T' = \boxed{\dfrac T2}

Hence, a + b = 1 + 2 = 3 a+b = 1+2 = 3 .

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Swati Tripathi
Oct 24, 2017

A= V 2 R \frac{V^2}{R}

V= 2 ∗ p i ∗ R T \frac{2*pi*R}{T}

A= 4 ∗ p i 2 ∗ R T 2 \frac{4*pi^2*R}{T^2}

8* 4 ∗ p i 2 ∗ R T 2 \frac{4*pi^2*R}{T^2} = 4 ∗ p i 2 ∗ 2 R t 2 \frac{4*pi^2*2R}{t^2} (t is new time period)

t=T/2

a=1,b=2 a+b=3

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