Physics- Circular motion question

[lDanny]

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Here is the problem:

There is object with mass of 30 kg that is swinging in a vertical circle (r = 1.4), at a speed at 6 m/s. At an angle of 30 degrees. What is the magnitude of the resultant force?

Don't you just use Fc= (mv^2)/R ? In the answer key is says the answer is 8.2N. What am I doing wrong?

Thanks

 

[PiBond]

I think your answer key is wrong. I'm a little rusty on physics but isn't the resultant force just the net force causing the object to move around in the circle? Also known as the centripetal force? Here's what I get:

At 30 degrees your Net force directed toward the center is:

Fcentripetal = F(t) - mgcos(theta)

First, I solved for the tension at the bottom of the circle and got it to be about 1071N.

Equation used: F(t) - mg = mv^2/r

Therefore: 1071N - (30kg)(10m/s^2)cos(30)

=1071 - 259 = 820N

Final answer I got was 820N. I may be wrong but your answer key seems close to my number just a few digits off. Maybe the transcriber typed it in wrong?

Just my two cents

 

[Edison Smith]

mass of 30 kg ,angle of 30 degrees these two are enough to have the solution:
F1=mg/cos(30)
Then,the final result F=2*F1

 

[PiBond]

mass of 30 kg ,angle of 30 degrees these two are enough to have the solution:
F1=mg/cos(30)
Then,the final result F=2*F1

You need more than this to solve the problem. Radius and velocity have to be incorporated given that this is a circle motion question.

 

[evusq]

I have a question relating to the centripetal force and the pendulum.
To find the tension in the string of pendulum, and the the speed through equlibrium is given, wouldn't you set it up so that T + mv^2/r =mg, since centrepetal force is always towards the center? The answer points the centripetal force in the same direction as mg... am I missing something?

Thanks!

 

[PiBond]

At the bottom of the path the tension of the string will be:

T - mg = mv^2/r

Tension and the weight of the object are pointing in different directions. The centripetal force (caused by the string) is pointing towards the center of the circle and the weight is point away (towards the ground) so they have opposite signs. The only time they have the same sign for tension and mg is when you are at the top of the path in circular motion, which I believe is what you have represented in your equation.

Hope that helps

I have a question relating to the centripetal force and the pendulum.
To find the tension in the string of pendulum, and the the speed through equlibrium is given, wouldn't you set it up so that T + mv^2/r =mg, since centrepetal force is always towards the center? The answer says T- mg =mv^2/r, and has the centrepetal force pointing in the same direction as mg... am I missing something?

Thanks!

 

[BennieBlanco]

Here is the problem:

There is object with mass of 30 kg that is swinging in a vertical circle (r = 1.4), at a speed at 6 m/s. At an angle of 30 degrees. What is the magnitude of the resultant force?

Don't you just use Fc= (mv^2)/R ? In the answer key is says the answer is 8.2N. What am I doing wrong?

Thanks

1st step is clarity. When is the object traveling at 6 m/s? At the top? At the bottom? At 30 degrees above the horizontal? The speed isn't constant. 2nd, 1.4 m ? 1.4 ft? 1.4 angstroms?

Physics becomes very difficult when there is no clarity. To solve a problem you must at least accurately define what you are trying to figure out.

The centripetal force dictates the speed. Therefore if you know the speed / radius / mass then the force is easy to calculate. The tension IS the centripetal force, which again, dictates the speed.

 

[ezsanche]

1st step is clarity. When is the object traveling at 6 m/s? At the top? At the bottom? At 30 degrees above the horizontal? The speed isn't constant. 2nd, 1.4 m ? 1.4 ft? 1.4 angstroms?

Physics becomes very difficult when there is no clarity. To solve a problem you must at least accurately define what you are trying to figure out.

The centripetal force dictates the speed. Therefore if you know the speed / radius / mass then the force is easy to calculate. The tension IS the centripetal force, which again, dictates the speed.

bingo!

 

[EyEnStein 07]

Is there a diagram for this? (im pretty sure there is) If so can you provide it or even recreate it and attach so we can see where you went wrong?

 

[himself]

wtf?

 

[mzblue]

Pi bond, where did you get the equation. i mean tension is a force so i know T = mv^2/r but how did you know 30 degrees means it's at the bottom and that the weight needs to subtracted. i'm curious cuz i just finished this topic in physics and while i had to use the above equation, i didn't know at some point the weight can be subtracted. if there's some study aid u found this in, lemme know so i can buy it to supplement my knowledge. my physics class isn't as rigid as it can be.

At the bottom of the path the tension of the string will be:

T - mg = mv^2/r

Tension and the weight of the object are pointing in different directions. The centripetal force (caused by the string) is pointing towards the center of the circle and the weight is point away (towards the ground) so they have opposite signs. The only time they have the same sign for tension and mg is when you are at the top of the path in circular motion, which I believe is what you have represented in your equation.

Hope that helps

 

[PiBond]

Pi bond, where did you get the equation. i mean tension is a force so i know T = mv^2/r but how did you know 30 degrees means it's at the bottom and that the weight needs to subtracted. i'm curious cuz i just finished this topic in physics and while i had to use the above equation, i didn't know at some point the weight can be subtracted. if there's some study aid u found this in, lemme know so i can buy it to supplement my knowledge. my physics class isn't as rigid as it can be.

The equation I used was determined by drawing a free-body diagram of an object in rotation at the bottom of its path (assuming it's a vertical path).

The object is being whirled around in a circle, therefore we know centripetal force is at work. This is the cue to use mv^2/r
At the bottom of the path there is the Tension of the string (upward direction) causing the object to accelerate in the circle--we can denote this T or F(t). There is also the force of gravity acting on the object (mg) in the downward direction. The force of gravity always points toward the center of the earth.

Conceptually, we can think of it in terms of where the most tension will be experienced in the path of rotation. This is at the bottom and can be proven mathematically if we solve for Tension at this point. T=mv^2/r + mg

After solving for tension at the bottom we know the tension at 30 degrees must be less. At 30 degrees our net forces acting on the object in rotation are mgcos30 and Tension, both pointing in opposite directions.