In: Physics
1) Rotational Dynamics: Four objects of equal mass start at the top of an inclined plane. A solid cylinder, a thin walled cylinder, a solid sphere, and a thin walled sphere. All objects start at rest. Starting at rest,
a. Write down the work energy theorem including both kinetic energy terms and all potential energy terms.
b. If all four objects are released at the same moment, how long does each take to reach the bottom of the incline plane?
c. Find the ratio of translational (linear) kinetic energy to rotational kinetic energy for each object.
d. Compare the results of the race obtained in part a to the results obtained in part b.
e. If the ramp is 20 m long and inclined at an angle od 20 degrees, what is the angular speed of the solid cylinder at the bottom of the ramp.
f. What torque is required to achieve this final angular speed?
given four objects of equal mass m
let radius of each object be r
moment of inertia
solid cylinder = mr^2/2
thin walled cylinder = mr^2
solid sphere = 2mr^2/5
hollow sphere = 2mr^2/3
a. let height of inline be h
then from work energy theorem
mgh = KE + RKE ( Where KE is the translational KE of the object and RKE is rotational KE of the object at the bottom of the incline)
b. time taken to reach the bottom = t
then
2*a*h/sin(theta) = v^2
v is speed at bottom of the incline
theta is angle of the incline
length of inclien = l
then
2a*l = v^2
and
v = at
t = v/a = sqrt(2l/a)
and 2*a*l = v^2
a = v^2/2l
t = v*2l/v^2 = 2l/v
for solid cylinder
mgh = 0.5mv^2 + 0.5*mv^2/2 = 0.75mv^2
v = sqrt(gh/0.75)
t = 2l*sqrt(0.75/gh)
for walled cylinder
mgh = 0.5mv^2 + 0.5mv^2 = mv^2
hence
t = 2l*sqrt(1/gh)
for solid spherer, mgh = 0.5mv^2 + 0.2mv^2
t = 2l*sqrt(0.7/gh)
for hollow sphere
mgh = 0.5mv^2 + 0.133mv^2
t = 2*l*sqrt(0.6333/gh)
c. TKE/RKE
for solid cylinder
m/0.5m = 2
for hollwo cylinder
m/m = 1
for solid sphere
m/(2m/5) = 2.5
for hollow sphere
m/(2m/3) = 1.5
d. the results from work energy theorem and the time taken to reach the bottom are in agreement with each other
e. l = 20 m
theta = 20 deg
h = l*sin(theta) = 20*sin(20) = 6.8404028 m
angular speed of solid cylinder at bopttom of ramp = w
t = 2l*sqrt(0.75/gh) = 4.22890635 s
0.5*mr^2*alpha = f*r ( f is forc e of friction)
alpha = 2f/mr
and
a = (mg*sin(theta) - f)/m = g*sin(theta) - r*alpha/2
also
a = alpha*r
hence
alpha*r*1.5 = g*sin(theta)
alpha = w/t
w*r*1.5/4.22890 = g*sin(20)
w = 9.459/r rad/s/s where r is radius of the cylinder
f. torque = f = alpha*mr/2