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# A1.50-kg grinding wheel is in the form of a solid cylinder ISBN: 9780321675460 31

## Solution for problem 33E Chapter 10

University Physics | 13th Edition

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Problem 33E

A1.50-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s? (b) Through what angle has it turned during that time? (c) Use Eq. (10.21) to calculate the work done by the torque. (d) What is the grinding wheel’s kinetic energy when it is rotating at 1200 rev/min? Compare your answer to the result in part (c).

Step-by-Step Solution:

Solution 33E Step 1 of 8: Given solid cylinder of radius R= 0.1 m with mass M= 1.5 kg rotates from rest ( = 0) i to angular speed =1200 rev/min in time t= 2.5 s. We need to calculate the torque f during this time, angle it has covered , work done by torque and finally have to compare this energy with kinetic energy due to rotation with speed 1200 rev/min. Given data, Radius, R= 0.1 m Mass , M=1.5 kg Time taken t= 2.5 s Initial angular speed, = 0 i Final speed, =1200 rev/min f Using 1 rev =2 rad and 1 min = 60 s = 1200 ×2rev/min f 60 = 125.6 rad/s f To find, (a) orque, = (b) Angular displacement, = (c) Work done by torque, W tor= (d) inetic energy if wheel , KE= Step 2 of 8: (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s In order to calculate the torque on the solid cylinder, we need to calculate moment of inertia and angular acceleration , To calculate the moment of inertia of the solid cylinder of mass M=1.5 kg and radius R=0.1 m Moment of inertia of the solid cylinder is given by, 1 2 I= M2 Substituting M=1.5 kg and R=0.1 m 2 I= (2.5 kg)(0.1) I = 0.0075 kg .m 2 Step 3 of 8: To calculate the angular acceleration during time t= 2.5 s Using equation of rotational kinematics, = t f i Where is angular acceleration. fi Solving for , = t Substituting t= 2.5 s, = 0 and = 125.6 rad/s i f 125.6 rad/s = 2.5 s = 50.24 rad/s 2 Step 4 of 8: To calculate the torque, Using the fundamental relation, = I Substituting I = 0.0075 kg .m and = 50.24 rad/s 2 = (0.0075 kg .m )(50.24 rad/s ) 2 = 0.377 N.m Therefore, the torque on the grinding wheel is 0.377 N.m.

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##### ISBN: 9780321675460

This full solution covers the following key subjects: min, wheel, torque, rev, grinding. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. The answer to “A1.50-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s? (b) Through what angle has it turned during that time? (c) Use Eq. (10.21) to calculate the work done by the torque. (d) What is the grinding wheel’s kinetic energy when it is rotating at 1200 rev/min? Compare your answer to the result in part (c).” is broken down into a number of easy to follow steps, and 80 words. Since the solution to 33E from 10 chapter was answered, more than 360 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. The full step-by-step solution to problem: 33E from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM.

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