Relating the Translational and Rotational Variables of a steam engine...?

Question

The flywheel of a steam engine runs with a constant rotational velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.2 h.

(a) What is the constant rotational acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many rotations does the wheel make during the slowdown?

(c) How many rotations does the wheel make before stopping?

(d) At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the translational acceleration of a flywheel particle that is 50 cm from the axis of rotation? (e) What is the magnitude of the net translational acceleration of the particle in (d)?

Can you explain your answers, and show the equations you used to solve this?

Answer 1

Wow. Your classmate just asked this question. But here's your answer anyways...

There are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.

(a) The initial energy of the system is:

E = (1/2) * I * w²

Where I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:

P = dE/dt = I * w * a

where a is the angular decceleration.

Since the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:

P = -E/t (negative sign denotes power loss)

Combining the power equations yields:

I * w * a = -E/t

Solving for a:

a = -E / (I * w * t)

Finally, substitute the expression for E to obtain:

a = -I * w² / (I * w * t)

a = -w/(t)

Substituting the values yields:

a = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)

a = -(15.71 rad/s)/(7920 s)

a = -.001983 rad/s²

(b) The rotational behavior of the fly wheel can be modeled as simply:

R(t) = -1/2*a*t² + w*t + Ri

where R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:

R(t) = -1/2*a*t² + w*t (radians)

R(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)

Subtituting the values:

R(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)

R = 9900 revolutions

(c) The tangential component of the linear acceleration of a point on a rotating body is given as:

A-tangent = a * r

where A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:

A-tangent = (-.001983 rad/s²)(.5 m)

A-tangent = -9.915 x 10^(-4) m/s²

A-tangent = -0.9915 mm/s²

(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:

Anet = [ (A-tangent)² + (A-normal)² ]^(1/2)

Or:

Anet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)

Note that the tangential velocity is given by:

v-tangent = w * r

Thus, the magnitude of net linear acceleration is:

Anet = [ (a * r)² + (w² * r)² ]^(1/2)

Substituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):

Anet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)

Anet = 30.81 m/s²

FINAL ANSWERS:

(a) a = -.001983 rad/s²
(b) R = 9900 revolutions
(c) A-tangent = -0.9915 mm/s²
(d) Anet = 30.81 m/s²

Answer 2

I answered the other question also, and my answer is different:

The starting angular velocity is 300 rpm. The wheel stops in 2.5 h. The acceleration (if constant) is 300 rpm / 2.5 h. You need the answer in revs / min^2, so you must convert 2.5 h to minutes (multiply by 60), then divide and get your answer.

The average rotational speed is (1/2)*300 rpm. Therefore in 2.5 h (2.5*60 min) it will make 150*2.5*60 revolutions.

Tangential acceleration aT = r*a, where a is the angular acceleration that you calculated in part (a). r is given as 50cm. Radial acceleration aR = w^2*r where w = angular velocity (given as 75rpm) and r = 50cm. The net linear acceleration is the vector sum of radial and tangential acceleration √[aT^2 + aR^2]

Remember to keep you units consistent (hours, mins, secs)

I don't give numerical results. You can figure it out.