Question

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.6 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 70.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 46 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answer 1

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 = -(140 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.6 hr * 3600 sec/hr)

a = -(14.66 rad/s)/(9360 s)

a = -.001566 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(2.6 hr) = R(9360 s) = (-.5*.001566*(9360 s)² + (14.66 rad/s)(9360 s))/(2*pi)

R = 32756.62 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=70 rev/min), the linear acceleration will also be constant. Thus:

A-tangent = (-.001566 rad/s²)(.46 m)

A-tangent = -7.2036 x 10^(-4) m/s²

A-tangent = -0.72036 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 70 rev/min is equal to 7.32667 rad/s):

Anet = [ (-.001566 * .46)² + (7.327² * .46)² ]^(1/2)

Anet = 24.69 m/s²

FINAL ANSWERS:

(a) a = -.001566 rad/s²
(b) R = 32756.62 revolutions
(c) A-tangent = -0.72036 mm/s²
(d) Anet = 24.69 m/s²