Guadalupe has a motorized globe on her desk that has a 0.16 m radius. She turns on the 4.25-watt motor and the globe begins to s
pin. The globe starts from rest and has a final rotational speed of 0.628 rad/s (1 revolution per 10 seconds). The globe reaches this speed in 25 seconds.
A. What is the angular displacement during the 25-second time interval in radians?
B. What is the linear speed of a point on the globe’s equator?
C. What is the average torque provided by the motor during the 25-second time interval?
A. What is the angular displacement during the 25-second time interval in radians?
B. What is the linear speed of a point on the globe’s equator?
C. What is the average torque provided by the motor during the 25-second time interval?
1 answer:
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Answer:
(a) coefficient of friction = 0.451
This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)
(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.
Explanation:
The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.
Thank you for reading and I hope this is helpful to you.
