Question

A 0.2 kg mass is rotated in a horizontal circle of radius 10 cm. If its angular speed is 0.4 rad/s, the centripetal force acting on it

Answer 1

Solution

• Force experienced by the particle is 32 × 10^-4 N

Given

• Mass of the particle,M = 0.2 Kg

• Radius of the circular path,r = 10cm = 10^-1 m

• Angular Velocity = 0.4 rad/s

Firstly,we would need to calculate linear velocity of the particle

Thus,

$\sf \: v = r \omega \\ \\ \hookrightarrow \sf \: v = {10}^{ - 1} \times 0.4 \\ \\ \hookrightarrow \ \sf v = \dfrac{4}{100} \\ \\ \hookrightarrow \: \boxed{ \boxed{ \sf v = 0.04 \: {ms}^{ - 1}}}$

Centripetal Force

$\tt \: F = \dfrac{ {mv}^{2} }{r} \\ \\ \longrightarrow \: \tt \: F = \frac{0.2 \times {0.04}^{2} }{ {10}^{ - 1} } \\ \\ \longrightarrow \tt \: F = 0.2 \times 0.04 \times 0.4 \\ \\ \longrightarrow \: \boxed{ \boxed{ \tt \: F = 32 \times {10}^{ - 4} \ N}} \:$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• Mass of object (m) = 0.2 kg
• Radius of circle (r) = 10 cm = 0.1 m
• Angular Speed (ω) = 0.4 rad/s

_________________________

To Find :

• Centripetal Force

_________________________

Solution :

We have formula :

$\large{\boxed{\sf{V \: = \: r \omega}}} \\ \\ \implies {\sf{v \: = \: 0.1 \: \times \: 0.4}} \\ \\ \implies {\sf{v \: = \: 0.04}} \\ \\ \underline{\sf{\therefore \: Velocity \: is \: 0.04 \: ms^{-1}}}$

__________________________________

Also, we have formula for Centripetal Force :

$\large{\boxed{\sf{F \: = \: \dfrac{mv^2}{r}}}} \\ \\ \implies {\sf{F \: = \: \dfrac{0.2 \: \times \: (0.04)^2}{0.1}}} \\ \\ \implies {\sf{F \: = \: \dfrac{0.2 \: \times \: 0.0016}{0.1}}} \\ \\ \implies {\sf{F \: = \: \dfrac{0.0032}{0.1}}} \\ \\ \implies {\sf{F \: = \: 0.032}} \\ \\ \underline{\sf{\therefore \: Centripetal \: force \: is \: 32 \: \times \: 10^{-4} \: N}}$