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### Course: Class 11 Physics (India)>Unit 9

Lesson 13: Centripetal force problem solving

# Mass swinging in a horizontal circle

In this video David explains how to solve for the tension and speed of a mass swinging at an angle in a horizontal circle. Created by David SantoPietro.

## Want to join the conversation?

• Why did we need to solve for T? Couldn't we just do Tx = Ty(tan30)?
• It is conceptually more difficult to do it the way you have suggested, although it does work as well. The end goal is to find the velocity of the mass. The way you have proposed IS NOT really a shortcut as well as you have to sub in mg for Ty and then when you get Tx from the equation, you equate that to Fc which gives you velocity at the end. In fact both ways are essentially the same, but remember, part (a) was to find the total tension. Your way finds Tx straight away and you have to another step to find the total tension. Sorry for righting so much.
• At the end of the video () , the speed you calculated was 2.38ms, yet when I did this in my calculator I got 4.7m, I checked to see if I did it wrong but I don't think I have.
• Hi! You'll want to make sure your calculator mode is set to degrees and not radians. I yielded the answer 2.38 m/s in degree mode and 4.7 m/s in radian mode. Hope this helps!
• Would it make a difference if there was a rope connecting the ball to the pole at a 90degree angle? What would the tension of both of the ropes be?
• Interesting idea. The problem is solved already so the extra rope could have zero tension not affecting the previous in any way. Additionally letting it take horizontal tension would allow the speed v to increase indefinitely, taking all extra centrifugal force and preventing the rope from rising (angle increasing). At lower speeds the second rope would be slack and not participate and the angle would reduce.
• Why is there no centripetal force if its moving in a circular motion?
• Good question! You are correct, there must be a centripetal force. At in the video, he writes down Newton's 2nd Law in the x-direction, which is the direction that is toward the center since the circle is horizontal. So we see that the centripetal force in this case is the horizontal component of the tension, Tx = Tsin(30). That is the only force in the horizontal plane, so that is equal to the mass times the centripetal acceleration, a = v²/r
• How would you solve for speed if you were not given mass?
• You actually don’t need the mass to solve for speed. For example, we can leave T in terms of mass (m), so that T = m*9.8 / cos30. Then, plugging this T into the derived equation for v [v = sqroot (Tsin30*Lsin30) / m], you get v = sqroot (m*9.8*sin30*Lsin30 / m*cos30). There is an m in the numerator and the m in the denominator, so they cancel.
• how do you find the period
• I'm assuming you mean period of time? If so: We know that time is equal to Distance divided by Speed. We know the distance the ball traveled (it's just the circumference of the circle), and we solved for the speed in this video. Simply plugging in those numbers into the formula yields your time. Hope this answers your question!
• Shouldn't the ball move like a pendulum? How is it being kept up? Is it because this an ideal situation with no air resistance?
• If the ball's velocity were smaller, then you would be correct - it would just act like a pendulum or slowly spiral downwards.

At around , he explains how we avoid that. We set the constraint that the ball has to stay on the horizontal circle. This means that there is no net force pulling the ball up or down, so vertical acceleration must be zero.

Essentially, the Y-component of the tension in the string perfectly cancels the downwards force of gravity, leaving the ball on the same horizontal circle.
• At - , why is Tx positive? If gravity is negative for pulling down in the y direction, why is that x is not negative? If it's going inward and to the left (opposite of positive x) should it be negative? Or am I missing something?
• Which direction you choose your x-axis or y-axis to be positive depends on you. I am sure you knew this already. Though the most important point here and the answer to your question is that you can apply Newton's 2nd law along y-axis (or, x-axis or whatever coordinate system you choose) independent of the direction in which positive x-axis lies.
Meaning, if you choose negative Y axis to be downward so that your force of gravity has a minus sign in the net force equation then for writing the net force equation along horizontal axis, you don't have to choose your +X axis to be to the right.
a_x and a_y along horizontal and vertical axes, respectively, are independent of each other. So the choice of positive axes along orthogonal coordinates axes is entirely up to you.
• He said "The only other force [than gravity] on this mass is tension", but why wasn't the normal force included?
(1 vote)
• The normal force wasn't one of the forces because the normal force results from contact. The ball is not in contact with anything but the rope which is pulling the ball from flying elsewhere and therefore the tension force.

Hope this helped!👍