Calculating Tangential & Angular Velocity of a Block on a String

  • Thread starter Thread starter akatsafa
  • Start date Start date
  • Tags Tags
    Circular
Click For Summary
SUMMARY

The discussion focuses on calculating the tangential and angular velocity of a wooden block with a mass of 0.29 kg attached to a string, which completes 15 rotations in 23.5 seconds at a radius of 0.70 meters. The correct angular velocity (\(\omega\)) is determined to be 4.01 rad/s, calculated using the formula \(\omega = \frac{15 \text{ rev}}{23.5 \text{ s}} \times \frac{2\pi}{1 \text{ rev}}\). The tangential velocity (v) is then found to be 2.807 m/s using the equation \(v = r \omega\). The initial confusion arose from not accounting for multiple revolutions in the calculations.

PREREQUISITES
  • Understanding of circular motion concepts
  • Familiarity with angular velocity and tangential velocity equations
  • Basic knowledge of radians and revolutions
  • Ability to perform unit conversions in physics
NEXT STEPS
  • Study the relationship between linear and angular motion in physics
  • Learn about centripetal acceleration and its calculations
  • Explore the concept of rotational dynamics and torque
  • Practice problems involving circular motion and velocity calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of angular and tangential velocity calculations.

akatsafa
Messages
42
Reaction score
0
I haven't worked with circular motions yet but i have these questions for lab. Could someone please help?

A group of students observes that a wooden block o mass o.29kg on a string makes 15 rotations in 23.5 seconds at a radius of 0.70 meters. (a) the blocks tangential velocity is.. (b) the blocks angular velocity is..

I tried using the equation v= 2pi r/T but that's not working. With that I got 0.87m/s but it's not right. What equations should I use for these?
 
Physics news on Phys.org
Hi, I think I got it but do check my work.

First we calculate the [tex]\omega[/tex] and then we use it to find [tex]v[/tex].


[tex]\omega = \frac {15 \text{rev}}{23.5 \text{s}} \times \frac {2\pi}{1 \text{rev}} = 4.01 \text{rad/s}[/tex]

Using [tex]v=r \omega[/tex] we get [tex]v=2.807 \text{m/s}[/tex]
 
Last edited:
thank you...that's right. I wasn't accounting for the 15 revolutions. I was just using one revolution.
 

Similar threads

Replies
15
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
18
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K