Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

A boy is swinging a toy on a piece of string in a vertical circle. The toy has a mass of 150 g and the radius of the circle is 0.8 m. a) He swings the toy with a linear velocity of 2 m/s. Will the toy move in a circle? Explain your answer. b) Another boy swings the toy with a linear velocity of 3.5 m/s. Work out the tension in the string at the top of the circle, at the bottom of the circle and halfway between the top and the bottom of the circle.​

Sagot :

jay40891
At the top of the circular motion, both weight and tension provides for centripetal force.

By Newton’s Second Law,
Fnet = ma
mg + T = mv^2/r (since a = v^2/r and weight = mg)

For toy to continue moving in circle at the top,

T > 0
mv^2/r - mg > 0
v >root (gr)

Hence, minimum speed toy must have is 2.80 m/s. Since linear velocity is lower than the minimum linear velocity, the toy will not move in circular motion.

b) Tension at top = mv^2/r - mg
= (0.15)(3.5)^2/0.8 - (0.15)(9.81)
= 0.825 N

Tension at bottom = mv^2/r + mg
= (0.15)(3.5)^2/0.8 + (0.15)(9.81)
= 3.77 N

In the middle, only Tension provides for centripetal force. Hence,
Tension = mv^2/r
= (0.15)(3.5)^2/0.8
= 2.30 N
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.