Calculating Tensions in a Vertical Circular Motion Problem

In summary: Yep it looks spot on to me :smile:. Conservation of energy is by far the easiest approach to solving circular motion, many students overlook this method initially as they want to use all the new equations (toys) they've been given :wink:. They forget their basic principles, well that's what I find anyway...
  • #1
cy19861126
69
0

Homework Statement


A stuntman whose mass is 70 kg swings from the end of a 4.0-m-long rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions on the rope that are required to make him follow his circular path,(a) at the beginning of his motion, (b) at a height of 1.5 m above the bottom of the circular arc, and (c) at the bottom of his arc.


Homework Equations


F = ma
F = m * v^2/r


The Attempt at a Solution


a) Using the equation F = m*v^2/r, F is 0 because v is 0
b) First, I used the energy approach attempting to solve the problem:
mgh = 0.5mv^2 + mgh,
9.8 * 4 = 0.5 * v^2 + 9.8 * 1.5
v = 7m/s.

As I get velocity, I plugged the value into
F = m * v^2 / r
F = 70kg * 7^2 / 4
F = 857N

I do not think this is right, since the value is too large. Also, I found out that I can calculate the angle of the distance he travelled: arccos(2.5/4) = 0.89rad --> pi/2 - 0.89 rad = 0.68 rad, this is the angle which he travelled. But I do not know where to plug this in anywhere.
 
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  • #2
Why do you think this value is too large? Compare your tension with the weight of the stuntman; does the value seem more realistic now?
 
Last edited:
  • #3
Hootenanny said:
Why do you think this value is too large? Compare your tension with the weight of the stuntman; does the value seem more realistic now?
Are you saying I did this right? Yeah, now you've said it, I think it is realistic... Wow, I thought I didn't get this chapter at all
 
  • #4
cy19861126 said:
Are you saying I did this right? Yeah, now you've said it, I think it is realistic... Wow, I thought I didn't get this chapter at all
Yep it looks spot on to me :smile:. Conservation of energy is by far the easiest approach to solving circular motion, many students overlook this method initially as they want to use all the new equations (toys) they've been given :wink:. They forget their basic principles, well that's what I find anyway...
 

1. What is rotational motion?

Rotational motion is a type of motion in which an object rotates or spins around an axis. It is often described in terms of angular displacement, velocity, and acceleration.

2. What are some examples of rotational motion?

Some examples of rotational motion include the spinning of a top, the rotation of a wheel, the swinging of a pendulum, and the orbit of planets around the sun.

3. How is rotational motion different from linear motion?

Rotational motion involves circular or angular movement around an axis, while linear motion involves movement in a straight line. Additionally, rotational motion is described in terms of angular quantities, while linear motion is described in terms of linear quantities.

4. What is the relationship between torque and rotational motion?

Torque is a measure of the force that causes rotational motion. It is directly proportional to the angular acceleration of an object and is affected by the distance from the axis of rotation.

5. How can rotational motion be calculated or predicted?

Rotational motion can be calculated using equations such as torque = moment of inertia * angular acceleration and angular velocity = initial angular velocity + angular acceleration * time. It can also be predicted using principles of conservation of angular momentum and angular kinetic energy.

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