Consider an aluminum annular disk with an outer radius 63.2 mm and inner radius
ID: 1792204 • Letter: C
Question
Consider an aluminum annular disk with an outer radius 63.2 mm and inner radius 7.9 mm. The mass of the disk is 464grams (HINT: the parameters of the disk are the same as in Prediction 1-1).
The disk is allowed to rotate on a frictionless table with the rotation axis at its center. The disk has a small pulley rigidly mounted at the top concentrically. The pulley's radius is 13 mm, and the mass of the pulley is negligible. A string is wrapped around the pulley, and a hanging mass of 19.6 g is tied at the other end of the string. When the mass falls under gravity, it causes the stainless steel annular disk to rotate. Ignore the string's mass, and assume that the string's motion is frictionless.
I need help with the last two questions. Please show your work.
What is the angular acceleration of the aluminum disk? 2.6 rad/s2 What is the angular speed of the aluminum disk 5.2 seconds after the hanging mass is released from rest? 113.52 rad/s At what linear speed is the hanging mass falling at this time? 0.18 How far away from its initial position is the falling mass at this time? (use kinematics to answer this part) 0.468 What is the kinetic energy of the falling mass at this time? 0.0003175 What is the rotational kinetic energy of the aluminum disk at this time? 0.08335 How much distance has the hanging mass been falling by this time? (use conservation of energy to answer this part) 0.4356Explanation / Answer
a) Angular Acceleration:-
Moment of Inertia, I = 0.5*M*(R2^2 - R1^2)
= 0.5*0.464*[ (63.2*10^-3)^2 - (7.9*10^-3)^2 ]
= 9.122*10^-4 kg/m^2
Torque T = m*g*R3
= 0.0196*9.8*(13*10^-3)
= 2.5*10^-3
So Angular Acceleration = T / I
= = 2.74 rad/s^2
b) Angular Speed w = Angular Accel * t
w = 2.74*5.2
w = 14.23 rad/s
c) v = r*w
= (13*10^-3)*14.23
= 0.185 m/s
e) Kinetic Eenrgy
K = 0.5mv^2
= 0.5*0.0196*0.185
= 0.000335368 J
f)
Rotatiional Energy
K' = 0.5 I w^2
= 0.5* (9.12*10^-4) *14.23^2
= 0.03376 J
g) K + K' = mgh
h = (K + K') / mg
= (0.000335368 + 0.03376) / (0.0196*9.8)
= 0.177 m