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## Class 11 Physics (India)

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

Lesson 13: Centripetal force problem solving

# Yo-yo in vertical circle example

In this video David explains how to find the tension in a string that is whirling a yo-yo in a vertical circle. Created by David SantoPietro.

## Want to join the conversation?

• please, how do we get (v^2)/r=a ? thanks
• () I don't understand why we include gravitational force as a centripetal forces because it is not always pointed at the centre of the circle. when the yoyo is at its lowest point, fg is pointed downwards.
• You are right. We don't always include gravitational force as contributing to the centripetal force. The example at is only about the case where the yoyo is at the highest point where gravity is pointed downwards and centripetally, so we include it.
When the yoyo is at its lowest point, gravitational force points downwards, which is opposite to the centripetal direction. So we have to subtract it from the tension to get the centripetal force.
When the yoyo is at the leftmost or rightmost point, we don't include gravitational force because the gravitational force at that point is pointing tangentially to the circle, which is unrelated to the centripetal direction.
• What if the yo-yo is on the left and right points of the circle (let's say it's turning counter-clockwise)? How would I include gravity (acceleration) in the circular velocity?
• (at ) Since the force of gravity is always pointing downward (or pointing in vertical direction), it has no horizontal component if you're considering the points at 3 o'clock and 9 o'clock. Therefore, you don't have to include the force of gravity. If you're considering other points on the left or right, you need to include the horizontal component of the force of gravity. You only include the horizontal component of a force when solving these problems because only forces with some horizontal direction can affect the circular motion of the object. Hope that helps!
• David gave examples in vertical circles. There is also a video on horizontal circle. But what if the yo-yo is swinging at another angle to the horizontal? For example, if the yo-yo has a mas 5g, the string length is 0.5m and gravity is 9.8m/s^2, and it is swung at an angle of 36 degrees to the horizontal (as opposed to 90 degrees when the yo-yo is swinging vertically) and accelerating at 7.9m/s in the direction it is moving, how do I figure out the tension?

Thanks,

Gor
• I think you should be able to figure out the tension force if you draw your force diagram, with the horizontal and vertical components of the tension force in their respective directions.
You should also specify the direction in which the Yo-yo is moving. Since you have said it is 'accelerating', I'm sure the Yo-yo is not climbing up the circle which it is tracing in space.

--HIH--

Ramana
• Why do we not include normal force to calculate net forces?
• Normal force is between two surfaces in contact. Given that there are no surfaces in contact here, normal force does not exist.
• Wouldn't the velocity at the bottom be greater than at the top due to conservation of energy? the ball is gaining kinetic energy while losing potential energy all the while still being pulled by the rope. This would change the calculation
• What if the Yo-yo string were pointing in a direction other than directly up or down, for example a slanting direction or directly horizontal to the left or right. What would the sign of Fg and Tension be in that case?
• Fg is always down and T is always in the direction of the string.
• If the yo-yo is at the bottom, why would mg still be included if Fc includes any forces that are pointing toward the center of the circle?
• Gravity doesn't go away, so at the bottom the force that is providing the centripetal acceleration needs to overcome it.