×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
A particle moves along a circle of radius R with a constant angular speed omega .Its displacement(only magnitude)in time t will be

```
2 years ago

Arun
25768 Points
```							Consider A and B to be the initial and final positions. Using simple trigonometric relations we can obtain the shortest distance or the Displacement between A and B by the following equation -AB = 2Rsin(theta/2)Here,R= radius of circletheta= the angular displacement = w*Tw= angular speedT= time takenSo, |AB| = 2Rsin(w*T/2)
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions