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furqannoor · Oct 19, 2009, 10:58 AM

For a mass of 1kg swinging in a virticle circle on a 1.5 m string. Find: a) the minimum tention neede to maintain a speed of 10 m/s at the top of the swing.

b) the tention needed to maintain a speed of 10m/s when the mass is at side of the of the swing.

Nhatkiem · Oct 19, 2009, 12:02 PM

Originally Posted by furqannoor

For a mass of 1kg swinging in a virticle circle on a 1.5 m string. find: a) the minimum tention neede to maintain a speed of 10 m/s at the top of the swing.

b) the tention needed to maintain a speed of 10m/s when the mass is at side of the of the swing.

In uniform circular motion, the speed is constant (not velocity, but its magnitude)

The first thing you would need to do is to create a position function for both the x and y axis with respect to time.

Next find $dx/dt$ , and $dy/dt$ for velocity (these should be set equal to 10 and 0 depending on your initial conditions! Think about the direction and components of the velocity vectors at the top and sides of the circle)

What were interested in however is acceleration to use in Tension=(net Mass)*(acceleration)

To find acceleration, you need to find $d^2x/dt^2$ and $d^2y/dt^2$
You may also see it as $dv_x/dt$ and $dv_y/dt$ .

Nhatkiem · Oct 19, 2009, 12:04 PM

Sorry, I just remembered, you won't need acceleration.

You can also find tension in circular motion using

$T=mv^2/r$

Much easier than finding the objects acceleration.

Nhatkiem · Oct 19, 2009, 12:18 PM

Originally Posted by furqannoor

For a mass of 1kg swinging in a virticle circle on a 1.5 m string. find: a) the minimum tention neede to maintain a speed of 10 m/s at the top of the swing.

b) the tention needed to maintain a speed of 10m/s when the mass is at side of the of the swing.

Sorry about all the confusion, I didn't realize it was a vertical circle.

Part A)

If you can imagine a weight at its position at the top of the circle, there are 2 forces pulling down on it. The force of interest (Tension) and the force of gravity. We can then say that at the top position, the net force will be

$F_n=F_T+F_g$

$F_g=m*g$
$F_n=mv^2/r$ , therefore

$F_T=mv^2/r-mg$

Think about how the forces chance as position changes. Tension always points in the direction of acceleration for uniform circular motion, however gravitational force is constant downwards with a value of mg.

ebaines · Oct 19, 2009, 12:27 PM

I'm going to guess that the OP is in high school, and consequently the calculus solution approach is not going to help.

The force that must act on a mass to cause it to move in a circle at uniform velocity is $F = mv^2/r$ , and this force must be directed toward the center of the circle. For this problem this force comes from two sources: the tension in the string and the weight of the mass due to gravity. Remember that forces are vectors, and so can be added and subtracted vectorally. So you have:

$<br /> \vec{F}\ = \ \vec {W} + \vec{T}<br />$

where $\vec {W}$ is force due to the object's weight and $\vec{T}$ is the tension in the string. Rearrange:

$<br /> \vec{T} = \vec {F} - \vec{W}<br />$

You have already been told the formula for the magnitude of $\vec{F}$ , and the magnitude of the object's weight is simply mg. You just need to remember to keep the directions of the forces straight. The magnitude of the total force on the object $\vec{F}$ is constant, but its direction varies - at the top of the arc $\vec{F}$ is pointed down, whereas at the bottom of the arc it is up (always toward the center of the circle); but $\vec{W}$ is always pointed down. Post back and tell us what answer you get for $\vec{T}$ .