In the problems below, use g = 9.8 m/s 2 NOTE: You WILL want to pop up a copy of

Question

In the problems below, use g = 9.8 m/s2

NOTE: You WILL want to pop up a copy of the manual (see above) to take this quiz!

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. 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/s25.2 m/s2    9.8 m/s29.8 rad/s230 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/s10 m/s    20 m/s35 rad/s60 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/s230 rad/s2    33 m/s233 rad/s255 m/s2

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

2.0 m/s24.4 m/s2    5.9 m/s26.6 m/s29.8 m/s2

e.) What is the tension on the string?

1.3 N

11 N    

12 N

13 N

20 N

f.) What is the speed of the mass when it reaches the floor?

3.6 m/s4.2 m/s    5.0 m/s5.4 m/s10 m/s

g.) What is the angular velocity of the disk at that point?

1.1 rad/s18 rad/s    21 rad/s25 rad/s27 rad/s

Explanation / Answer

a.) T= Ia

       15*1/5= 1/2 *5*1/25 a

      a= 30 rad/s^2

b) w= u+at= 30*2=60 rad/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

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