ZuoTi.Pro
Question

The 2.5 kg, 35-cm-diameter disk in the figure is spinning at 300 rpm. Part A) How...

The 2.5 kg, 35-cm-diameter disk in the figure is spinning at 300 rpm.

Part A) How much friction force must the brake apply to the rim to bring the disk to a halt in 2.50 s?

science   physics   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

Here ,

let the frictional force needed is f

change in angular momentum = Torque * time taken

moment of inertia * change in angular speed = friction * radius * time

0.50 * 2.5 * (0.35/2)^2 * (300 * 2pi/60) = F * (0.35/2) * 2.5

solving for

F = 2.75 N

the frictional force needed is 2.75 N

1 0
Add a comment
Know the answer?
Add Answer to:
The 2.5 kg, 35-cm-diameter disk in the figure is spinning at 300 rpm. Part A) How...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • The 3.20 kg , 31.0-cm-diameter disk in the figure (Figure 1) is spinning at 310 rpm...

    The 3.20 kg , 31.0-cm-diameter disk in the figure (Figure 1) is spinning at 310 rpm . How much friction force must the brake apply to the rim to bring the disk to a halt in 2.60 s ? Brake

  • The 3.50 kg. 35.0-cm-diameter disk in the figure (Figure 1lis spinning at 320 rpm. Part A...

    The 3.50 kg. 35.0-cm-diameter disk in the figure (Figure 1lis spinning at 320 rpm. Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.40 s? Express your answer with the appropriate units. HA + O ? Value Units Submit Request Answer Provide Feedback Figure < 1 of 1 > Brake

  • The 3.50 kg. 35.0-cm-diameter disk in the figure (Figure 1) spinning at 320 rpm. Part A...

    The 3.50 kg. 35.0-cm-diameter disk in the figure (Figure 1) spinning at 320 rpm. Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.40 s? Express your answer with the appropriate units. HÅ on ? 42.8 N Submit Previous Answers Request Answer X incorrect; Try Again, 8 attempts remaining Figure < 1 of 1 Provide Foodback Brake

  • Constants Part A A solid, uniform cylinder with mass 8.30 kg and diameter 19.0 cm is...

    Constants Part A A solid, uniform cylinder with mass 8.30 kg and diameter 19.0 cm is spinning with angular velocity 205 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction- brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.345. What must the applied normal force be to bring the cylinder to rest...

  • A rotor wheel disk has a mass of 2.0 kg and is 30 cm in diameter....

    A rotor wheel disk has a mass of 2.0 kg and is 30 cm in diameter. At the operating speed it rotates at a speed of 300 rpm. If only friction was applied from the brake pads at the edge of the disk, how much frictional force is need to stop the disk in 3.0 s?

  • A solid, uniform cylinder with mass 8.05kg and diameter 20.0cm is spinning with angular velocity 240rpm...

    A solid, uniform cylinder with mass 8.05kg and diameter 20.0cm is spinning with angular velocity 240rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction-brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.336. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.50 revolutions?

  • A piece of metal is pressed against the rim of a 3.4- kg, 24- cm-diameter grinding...

    A piece of metal is pressed against the rim of a 3.4- kg, 24- cm-diameter grinding wheel that is turning at 2800 rpm. The metal has a coefficient of friction of 0.70 with respect to the wheel. When the motor is cut off, with how much force must you press to stop the wheel in 20.0 s?

  • equipment. A 3.0-m-diameter merry-go-round with a mass of 300 kg is spinning at 15 rpm ....

    equipment. A 3.0-m-diameter merry-go-round with a mass of 300 kg is spinning at 15 rpm . John runs tangent to the merry-go-round at 4.8 m/s , in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. Part A: What is the merry-go-round's angular velocity, in rpm, after John jumps on?

  • , n Review Constants Part A A flywheel in a motor is spinning at 580 rpm...

    , n Review Constants Part A A flywheel in a motor is spinning at 580 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm. The power is off for 37.0 s, and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. At what rate is the flywheel spinning when the power comes back...

  • A 45 cm -diameter potter's wheel with a mass of 15 kg is spinning at 180...

    A 45 cm -diameter potter's wheel with a mass of 15 kg is spinning at 180 rpm. Using her hands, a potter forms a 14 cm-diameter pot that is centered on and attached to the wheel. The pot's mass is negligible compared to that of the wheel. As the pot spins, the potter's hands apply a net frictional force of 1.3 N to the edge of the pot. If the power goes out, so that the wheel's motor no longer...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT