Question

A particle perform uniform circular motion with radius 1 meter .If the frequency of revolution is 120 rpm Find -
1. period of revolution
2. linear speed
3. centripetal acceleration.​

Answer 1

Answer:

Hey mate

Explanation:

here is ur answer

Answer 2

Answer:-

$\red{\bigstar}$ Period of revolution = 0.5 sec

$\red{\bigstar}$ Linear speed = 12.56 m/s

$\red{\bigstar}$ Centripetal acceleration = 50.24 rad/s²

• Given:-

Radius [r] = 1 m

Revolutions = 120 rpm

• To Find:-

1. Period of revolution.

2. Linear speed

3. Centripetal acceleration

• Solution:-

120 rpm → rps

→ $\sf \dfrac{120}{60}$

→ 2 rps

$\sf\red{\omega = 2 \pi n}$

where n = no. of revolutions

→ $\sf{\omega = 2 \pi \times 2}$ rad/s

→ $\sf{\omega = 4 \pi }$ rad/s

• Case 1:-

$\pink{\bigstar}$ $\sf{\boxed{\red{Time \: period <strong>[</strong><strong>T</strong><strong>]</strong><strong> </strong><strong>=</strong><strong> </strong> \dfrac{2 \pi}{\omega}}}}$

T = $\sf \dfrac{2 \pi}{4 \pi}$

T = $\sf \dfrac{1}{2}$

T = 0.5 sec.

• Case 2:-

$\pink{\bigstar}$$\sf{\boxed{\red{Linear \: velocity[v] = \omega \times r}}}$

v = 4π × 1

v = 4π

v = 4 × 3.14

v = 12.56 m/s

• Case 3:-

$\pink{\bigstar}$$\sf{\boxed{\red{Centripetal \: acceleration [a] = r \omega ^{2}}}}$

a = 1 × (4π)²

a = 1 × 16π

a = 16π

a = 16 × 3.14

a = 50.24 rad/s²