Tension in Circular Motion

Tension in Circular Motion

Assessment

Interactive Video

Physics, Mathematics, Science

9th - 12th Grade

Hard

Created by

Patricia Brown

FREE Resource

The video tutorial explains how to find the tension at Point P by reviewing the velocity equation and deriving the tension equation. It covers the components of forces acting on the object, such as mg cos theta and centripetal force, and presents the tension equation in two forms: in terms of theta and without theta. The tutorial simplifies the tension equation and concludes with a preview of the next video, which will discuss tensions at various points.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the initial velocity equation mentioned in the video?

V^2 = U^2 - 2gh

V = U - 2gh

V = U + 2gh

V^2 = U^2 + 2gh

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which force acts downward on the object?

Gravitational force

Frictional force

Centrifugal force

Centripetal force

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the formula for centripetal force in this context?

mgh

mu^2/r

mv^2/r

mg cos(theta)

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the two components responsible for tension?

mv^2/r and centrifugal force

mg sin(theta) and mg cos(theta)

mg sin(theta) and mv^2/r

mg cos(theta) and mv^2/r

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the tension equation expressed in terms of velocity and height?

T = mgh + mu^2/r

T = mg cos(theta) + mu^2/r - 2mgh/r

T = mg sin(theta) + mu^2/r

T = mg cos(theta) - mu^2/r + 2mgh/r

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the expression for cos(theta) used in the tension equation?

H + R / r

H - R / r

R - H / r

R + H / r

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the final form of the tension equation in terms of theta?

T = m(g cos(theta) + u^2/r - 2g)

T = m(u^2/r - 3gh + rg)

T = m(g cos(theta) + 3g cos(theta) - 2g)

T = m(u^2/r + 2g cos(theta) - 3g)

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