What is the Speed at the Bottom of a Vertical Circle?

[WhiteWolf98]

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

 

[berkeman]

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.

 

[ehild]

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.

 

[WhiteWolf98]

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' ...

 

[berkeman]

http://www.schoolphysics.co.uk/age1...text/Motion_in_a_vertical_circle/images/1.png

 

[WhiteWolf98]

How is the velocity when the string is horizontal related to the velocity when the string is vertical (on the bottom)?

 

[berkeman]

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.

 

[WhiteWolf98]

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

 

[berkeman]

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...

 

[WhiteWolf98]

$$\frac 1 2 mv^2+mgh=\frac 1 2 mu^2 $$
$$mv^2+2mgh=mu^2$$
$$m(v^2+2gh)=mu^2$$
$$u=\sqrt {v^2+2gh}$$
$$Thank~you!$$