9. A 0.500 kg circular disc with a radius of 0.700 m is bound to rotate about its center. Three forces act on the object as depicted in the diagram below. Fj=5.00N, F=7.00 N and F3-5.00N. F3 is 0.400 m from the axis of rotation while the other two forces act tangentially. The moment of inertia the sheet about its center is I = mR2 35° F3 F1 F2 a. What is the net torque acting on the disc? b. If the object is initially at rest, what will its rotational kinetic energy be 7.00 s after the forces begin to be applied?

9. A 0.500 kg circular disc with a radius of 0.700 m is bound to rotate about its center. Three forces act on the object as depicted in the diagram below. Fj=5.00N, F=7.00 N and F3-5.00N. F3 is 0.400 m from the axis of rotation while the other two forces act tangentially. The moment of inertia the sheet about its center is I = mR2 35° F3 F1 F2 a. What is the net torque acting on the disc? b. If the object is initially at rest, what will its rotational kinetic energy be 7.00 s after the forces begin to be applied?

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter8: Rotational Motion

Section: Chapter Questions

Problem 93A

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Concept explainers

Angular speed, acceleration and displacement

Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.

Angular Position

Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.

Question

9. A 0.500 kg circular disc with a radius of 0.700 m is bound to rotate about its center. Three
forces act on the object as depicted in the diagram below. F1=5.00 N, F=7.00 N and F3-5.00 N.
F3 is 0.400 m from the axis of rotation while the other two forces act tangentially. The moment
of inertia the sheet about its center is I ==mR2
%3D
1.
35°
F3
FI
F2
a. What is the net torque acting on the disc?
b. If the object is initially at rest, what will its rotational kinetic energy be 7.00 s after the forces
begin to be applied?

Expert Solution

Step 1

Given that:-

Mass of disc=0.50kg

Radius of disc=0.7m

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