Question

the distance of a particle moving on a circle of radius 12 m measured from a fixed point on the circle and measured along the circle is given by s = 2t3 (in metres). find the ratio of its tangential to centripetal acceleration at t = 2s.

Answer 1

Given,

S = 2t³

Differentiating both the sides with respect to the time.

dS/dt = d(2t³)/dt

⇒ Velocity = 2 × 3 × t²

⇒ Velocity = 6t².

Centripetal Acceleration = V²/R = (6t²)²/12 = 36t⁴/12 = 3t⁴  = 48 m/s².

Now, Tangential Acceleration = d|v|/dt = d|6t²|/dt = 2 × 6 × t = 12t  = 12 × 2 = 24 m/s².

∴ Ratio of Tangential and Centripetal acceleration = 12t/3t⁴

= 24/48

= 1/2

= 1 : 2

Hope it helps.

Answer 2

Centripetal acceleration of the particle is given by,

ac = v^2/r = v^2/12 -- (1)

and distance measured s = a t^3

ds/dt = v = 3 a t^2 and d^2s/dt^2  = at = 6at

The ratio of ac to at is given by,

ac/at = v^2/12/6at = 3at^2/72at = t/24

Therefore ratio of these two acceleration at t = 2s is given by,

ac/at = 2/24 = 1/12 or 1:12