Just as you had done in your lab, imagine that you have a solid uniform disk (sa

Question

Just as you had done in your lab, imagine that you have a solid uniform disk (same as a uniform cylinder) with a string wound around its rim. (These disks are truly solid and do not have an inner radius like those in the lab.) You pull the string with a force having a magnitude 15 N. The disk has a mass of 5 kg and a radius of 20 cm. The disk is free to rotate without friction about a fixed axis going through its center..

a.) What is the angular acceleration of the disk? 3.0 m/s2 5.2 m/s2 9.8 m/s2 9.8 rad/s2 30 rad/s2

b.) Suppose that you start pulling the string when the disk was at rest. What is the angular velocity of the disk after 2 s?

6.0 m/s 10 m/s 20 m/s 35 rad/s 60 rad/s Now, instead of pulling the string, you decide to attach a 2 kg mass to the free end of the string, hold it temporarily, and then release it (just as in the lab) from a height 1.5 m above the floor.

Take the positive direction to be pointing downward.

c.) What is the angular acceleration of the disk? Hint: look at the equation for angular acceleration in the lab manual.

22 rad/s2 30 rad/s2 33 m/s2 33 rad/s2 55 m/s2

d.) What is the acceleration of the falling mass?

2.0 m/s2 4.4 m/s2 5.9 m/s2 6.6 m/s2 9.8 m/s2

e.) What is the tension on the string?

1.3 N 11 N 12 N 13 N 20 N

Explanation / Answer

a) torque= 15*20/100 = 3 Nm

torque = I*a

3 = mr^2/2 * a

=> 3 = 5*(.2*.2)/2 *a

=> a = 30 rad/s^2 answer

b) v = at => v = 30*2 = 60 rad/sec

c) T*r = mr^2/2 *a

=> T*20/100 = 5*.2*.2/2 *a

Mg-T = Mra

=> T = Mg -Mra

=>(Mg -Mra)*r = mr^2/2 *a

=> 2*(9.8-(.2*a))*.2 = 5*.2*.2/2 *a

=> 2*9.8*.2 - .2*.2*2*a = 5*.2*.2/2*a

=> a = 21.7777778 = 22 rad/sec

d) A = ra = .2*22 = 4.4 m/s

e)T = (2*9.8) - (2*.2*21.7777778) = 10.88888888 = 11 N

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