Question

0.2 kg ball rotates at a constant of 3m/s on the end of 1.2 m long string. The string describes a horizontal circle. What is the centripetal acceleration of the object ? Please help me

Answer 1

Answer:

7.5 m/s^2

Explanation:

centripetal acceleration= v^2/r = 3^2/1.2= 9/1.2= 7.5

Answer 2

Given :

• Mass of the ball = 0.2 kg

• Length of the string = Radius = 1.2 m

• Velocity = 3 m/s

To Find :

• Centripetal Acceleration of the ball

Solution :

Centripetal Acceleration of a body is given by ,

$\boxed{ \rm{a = \frac{ {v}^{2} }{r} }}$

Where ,

• a is centripetal acceleration
• v is velocity of the body
• r is radius

We have ,

• v = 3 m/s
• r = 1.2 m

$: \implies \rm \: a = \dfrac{(3 {}^{} \: m {s}^{ - 1}) {}^{2} }{1.2 \ m} \\ \\ : \implies \rm \: a = \frac{9 \ \cancel{m {}^{2} }s {}^{ - 2} }{1.2 \ \cancel{m}} \\ \\ : \implies \rm a = \frac{9 \ m {s}^{ - 2} }{1.2} \\ \\ : \implies {\boxed {\rm {\blue{a = 7.5 \: m {s}^{ - 2} }}}}$

∴ The Centripetal acceleration of the ball is 7.5 ms⁻²

Additional Info :

• Centripetal acceleration is the amount of acceleration required to keep the body in circular motion

• Centripetal acceleration always acts towards the centre

• SI unit of Centripetal acceleration is ms⁻²