Circular Motion Speed Question

In summary, a 5 kg mass is performing circular motion at the end of a 3 m light, inextensible string. The speed of the mass is 2 m/s when the string is horizontal and the question asks for its speed at the bottom of the circle. The answer is 8 m/s and the problem should have stated that the circle is vertical. The relevant equations are T = mv^2/r and T = mu^2/r + mg, where T is tension, m is mass, v is velocity, r is radius, and g is gravity. Using the Conservation of Energy equation, we can find the speed at the bottom of the circle to be u = √(v^2 + 2gh
  • #1
WhiteWolf98
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5

Homework Statement


A ##5 ~kg## mass performs circular motion at the end of a light, inextensible string of length ##3~m##. If the speed of the mass is ##2 ~ms^{-1} ## when the string is horizontal, what is its speed at the bottom of the circle?

(assume ##g=10~ms^{-1}##) (Ans: ##8~ms^{-1}##)

I'm not sure in what plane? as such, to imagine this question. I know the equations for a horizontal and vertical circle, as well as a conical pendulum, but the question just doesn't make sense to me. Any help would be appreciated. Thank you

Homework Equations

The Attempt at a Solution

 
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  • #2
WhiteWolf98 said:

Homework Statement


A ##5 ~kg## mass performs circular motion at the end of a light, inextensible string of length ##3~m##. If the speed of the mass is ##2 ~ms^{-1} ## when the string is horizontal, what is its speed at the bottom of the circle?

(assume ##g=10~ms^{-1}##) (Ans: ##8~ms^{-1}##)

I'm not sure in what plane? as such, to imagine this question. I know the equations for a horizontal and vertical circle, as well as a conical pendulum, but the question just doesn't make sense to me. Any help would be appreciated. Thank you

Homework Equations

The Attempt at a Solution

Sure sounds like it's a vertical circle. Is there a figure with the problem or earlier problems? And please list the Relevant Equations and start working through them. Thanks.
 
  • #3
WhiteWolf98 said:

Homework Statement


A ##5 ~kg## mass performs circular motion at the end of a light, inextensible string of length ##3~m##. If the speed of the mass is ##2 ~ms^{-1} ## when the string is horizontal, what is its speed at the bottom of the circle?

(assume ##g=10~ms^{-1}##) (Ans: ##8~ms^{-1}##)

I'm not sure in what plane? as such, to imagine this question. I know the equations for a horizontal and vertical circle, as well as a conical pendulum, but the question just doesn't make sense to me. Any help would be appreciated. Thank you

Homework Equations

The Attempt at a Solution

You are right, the problem should have state the orientation of the circle. From the answer given, it should be vertical.
 
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  • #4
There were two questions before this that both involved a conical pendulum.

I thought it was a vertical circle too, but the only equations I know related to a vertical circle are to do with the top and bottom of the circle. I don't understand the part when it says, 'the string is horizontal' ...
 
  • #5
http://www.schoolphysics.co.uk/age1...text/Motion_in_a_vertical_circle/images/1.png
1.png
 

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  • #6
How is the velocity when the string is horizontal related to the velocity when the string is vertical (on the bottom)?
 
  • #7
WhiteWolf98 said:
How is the velocity when the string is horizontal related to the velocity when the string is vertical (on the bottom)?
Is it greater at the bottom. For more detail, you need to do the calculations. :smile:
 
  • #8
If the tension at the point where the string is horizontal is ##T_1## and the velocity is ##v##,
and the tension at the point where the string is vertical is ##T_2## and the velocity is ##u##, then we get the two equations:

$$T_1=\frac {mv^2} r$$

$$T_2=\frac {mu^2} r+mg$$

I don't know what to do next
 
  • #9
WhiteWolf98 said:
If the tension at the point where the string is horizontal is ##T_1## and the velocity is ##v##,
and the tension at the point where the string is vertical is ##T_2## and the velocity is ##u##, then we get the two equations:

$$T_1=\frac {mv^2} r$$

$$T_2=\frac {mu^2} r+mg$$

I don't know what to do next
Hint -- Using the Conservation of Energy can sometimes help to solve these problems...
 
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  • #10
$$\frac 1 2 mv^2+mgh=\frac 1 2 mu^2 $$
$$mv^2+2mgh=mu^2$$
$$m(v^2+2gh)=mu^2$$
$$u=\sqrt {v^2+2gh}$$
$$Thank~you!$$
 

1. What is circular motion speed?

Circular motion speed is the speed at which an object moves around a circular path. It is a measure of how fast an object is moving as it travels along the circumference of a circle.

2. How is circular motion speed calculated?

Circular motion speed can be calculated by dividing the distance traveled around the circle by the time it takes to complete one full revolution. It can also be calculated by multiplying the radius of the circle by the angular velocity (rate of change of angle).

3. What is the difference between linear speed and circular motion speed?

Linear speed is the distance traveled per unit of time in a straight line, while circular motion speed is the distance traveled per unit of time around a curved path. Linear speed is constant, but circular motion speed is constantly changing due to the change in direction of the object.

4. How does the radius of the circle affect the circular motion speed?

The larger the radius of the circle, the higher the circular motion speed. This is because the object has to travel a greater distance in the same amount of time, resulting in a higher speed. Conversely, a smaller radius will result in a lower circular motion speed.

5. Can circular motion speed be negative?

Yes, circular motion speed can be negative if the object is moving in a clockwise direction. This indicates that the object is moving in the opposite direction of the positive angular velocity. However, the magnitude of the speed will still be the same regardless of the direction of movement.

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