1) Rotational Dynamics: Four objects of equal mass start at the top of an inclin

Question

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?

Explanation / Answer

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

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