1. A grindstone spinning at a rate of 20 rev/s has what approximate angular spee

Question

1. A grindstone spinning at a rate of 20 rev/s has what approximate angular speed?

2 A 0.50 kg mass is whirled around at the end of a 1.00 m string. The mass sweeps out a horizontal circle. The string has a 500 N capacity, what is the maximum tangential speed of the mass if the string doesn't break?

3. A 1,000 kg car rounds an unbanked curve with a radius of 50 m at a speed of 10 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping ?

4. An object of mass 1.00 kg is transported to the surface of Planet Z, where an object's weight is measured to be 5 N. The radius of the planet is 5.0x10^4 m. What is the mass of Planet Z?

5. A 0.5 m radius grinding wheel takes 6 s to speed up from 2.0 rad/s to 12.0 rad/s. What is the wheel's average angular acceleration?

6. A planet has 2 times the radius of the Earth, but has the same density as the Earth, What is the gravitational acceleration at the surface of the planet? (use g = 9.8 m/s^2)

7. If the net torque is applied to an object, that object will experience a non zero:

8. A disk has a moment of inertia of 3.0 x 10^-4 kg*m^2 and rotates with an angular speed of 5.0 rad/sec. What net torque must be applied to bring it to rest within 2.0 s?

9. A bucket filled with water has a mass of 5.0 kg and is attached to a rope which in turn is wound around a 0.050 m radius cylinder at the top of a well. What torque does the weight of wate rand bucket produce on the cylinder if the cylinder is not permitted to rotate?

10. A bowling ball has a mass of 7.0 kg, a moment of inertia of 2.8 x 10^-2 kg*m^2 and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 5.0 m/s, what is its total kinetic energy?

Explanation / Answer

1)angular speed=20*2pi

=125.66 rad/s

2)mv^2/r=T

or 0.5*v^2/1=500

or v=31.62 m/s

3)umg=mv^2/r

or u=v^2/rg

=0.204

5)angular acceleration=(12-2)/6

=10/6

=1.667 rad/s^2

6)g=9.8*2

=19.6 m/s^2

7) angular acceleration

8)alpha=5/2

=2.5 rad/s^2

so torque=I*alpha

=2.5*3*10^-4

=7.5*10^-4 Nm

9)torque=mgl

=5*9.8*0.05

=2.45 Nm

10)kinetic energy=0.5mv^2+0.5*Iv^2/r^2

=0.5*7*5*5+0.5*2.8*10^-2*5*5/(0.01)

=122.5 J

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