cq_1_261

phy 201

Your 'cq_1_26.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

A simple pendulum has length 2 meters. It is pulled back 10 cm from its equilibrium position and released. The tension in the string is 5 Newtons.

• Sketch the system with the pendulum mass at the origin and the x axis horizontal.

answer/question/discussion:

• Sketch a vector representing the direction of the pendulum string at this instant. As measured from a horizontal x axis, what is the direction of this vector? (Hint: The y component of this vector is practically the same as the length; you are given distance of the pullback in the x direction. So you know the x and y components of the vector.)

answer/question/discussion:

down and slightly to the left of the negative y axis. The vector is in the 3rd quadrant.

• What is the direction of the tension force exerted on the mass?

answer/question/discussion:

Since the force pulling the mass back is to the left, in the 180 degree direction, the tension force is counter-acting, pointing in the 0 degree direction.

The tension in the string must be in the direction of the string; it can't be at 0 deg or 180 deg. A string can only exert a force in its direction.

The string does counter the pullback force, but it also holds the mass up against gravity.

• What therefore are the horizontal and vertical components of the tension?

answer/question/discussion:

Therefore, the horizontal and vertical components of the tension are in the opposite direction of the normal forces of gravity and pulling the mass back. The resultant would be in the first quadrant, somewhere slightly to the right of 90 degrees.

• What therefore is the weight of the pendulum, and what it its mass?

answer/question/discussion:

not enough information given. I have a lot of stuff written here in front of me but it’s too much to type out in a word document. Basically the angle of the mass’ net force is at 267 degrees and the answers can be expressed in terms of F=m(v^2/r).

The gravitational force is at 267 deg with respect to the direction of the pendulum. So you're definitely on the right track.

A centripetal force equal to m v^2 / r is relevant when the pendulum is in motion, but in this situation it's not in motion

• What is its acceleration at this instant?

answer/question/discussion:

30 minutes

You've got most of it.

The bottom line is that the horizontal component of the tension provides the restoring force, and the vertical component keeps the mass from falling vertically.

&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.

Solution

This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#

cq_1_261

Your 'cq_1_26.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** **

The tension in the string must be in the direction of the string; it can't be at 0 deg or 180 deg. A string can only exert a force in its direction. The string does counter the pullback force, but it also holds the mass up against gravity.

The gravitational force is at 267 deg with respect to the direction of the pendulum. So you're definitely on the right track. A centripetal force equal to m v^2 / r is relevant when the pendulum is in motion, but in this situation it's not in motion

** **

** **

You've got most of it. The bottom line is that the horizontal component of the tension provides the restoring force, and the vertical component keeps the mass from falling vertically.

The tension in the string must be in the direction of the string; it can't be at 0 deg or 180 deg. A string can only exert a force in its direction.

The string does counter the pullback force, but it also holds the mass up against gravity.

The gravitational force is at 267 deg with respect to the direction of the pendulum. So you're definitely on the right track.

A centripetal force equal to m v^2 / r is relevant when the pendulum is in motion, but in this situation it's not in motion

You've got most of it.

The bottom line is that the horizontal component of the tension provides the restoring force, and the vertical component keeps the mass from falling vertically.