A 1.60kg grinding wheel is in the form of a solid cylinder of radius 0.200m . Pa

Question

A 1.60kg grinding wheel is in the form of a solid cylinder of radius 0.200m .

Part A

What constant torque will bring it from rest to an angular speed of 1500rev/min in 2.40s ?

Part B

Through what angle has it turned during that time?

Part C

Use equation W=?z(?2??1)=?z?? to calculate the work done by the torque.

Part D

What is the grinding wheel's kinetic energy when it is rotating at 1500rev/min ?

Part E

Compare your answer in part (D) to the result in part (C).

Explanation / Answer

w = 1400rpm = 1500*2*3.14/60 = 157 rad/s

alfa = w/t = 157/2.4 = 65.417

partA)

torque = I*alfa = 0.5*M*R^2*alfa = 0.5*1.6*0.2*0.2*65.417 = 2.093344 Nm

++

part B)

theta = 0.5*alfa*t^2 = 0.5*65.417 *2.4*2.4 = 188.40096 radians = 30 revolutions

++

patr C

W = 2.093344*188.40096 = 394.4 J

part D)

KE = 0.5*I*w^2 = 0.5*0.5*1.6*0.2*0.2*157*157 = 394.4 J

E) The results are same

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.