Solution Found!
Solved: A1.50-kg grinding wheel is in the form of a solid
Chapter 10, Problem 33E(choose chapter or problem)
A \(\text { 1.50-kg }\) grinding wheel is in the form of a solid cylinder of radius
\(0.100 \mathrm{~m}\).
(a) What constant torque will bring it from rest to an angular speed of in \(2.5 \mathrm{~s}\)?
(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 Compare your answer to the result in part (c).
Equation Transcription:
Text Transcription:
1.50-kg
0.100 m
2.5 s
Questions & Answers
QUESTION:
A \(\text { 1.50-kg }\) grinding wheel is in the form of a solid cylinder of radius
\(0.100 \mathrm{~m}\).
(a) What constant torque will bring it from rest to an angular speed of in \(2.5 \mathrm{~s}\)?
(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 Compare your answer to the result in part (c).
Equation Transcription:
Text Transcription:
1.50-kg
0.100 m
2.5 s
ANSWER:
Solution 33E
Step 1 of 8:
Given solid cylinder of radius R= 0.1 m with mass M= 1.5 kg rotates from rest () to angular speed
1200 rev/min in time t= 2.5 s. We need to calculate the torque 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,
-
Final speed,
1200 rev/min
Using 1 rev =2rad and 1 min = 60 s
To find,
(a)Torque, =?
(b) Angular displacement, =?
(c) Work done by torque,
(d) Kinetic energy if wheel , KE=?