The flywheel of a steam engine runs with a constant angular velocity of 220 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.7 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 110 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 47 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?
Please show the equations and solutions especially for c and d.
I need to see your work or thinking to critique it. On c, tangential component is wr. Net linear acceleration will be the vector addition of tangential acceleration and centripetal acceleration.
c) w = 110 rev/min
a = (alpha)r
What equation do I use for getting alpha? Is the angular acceleration constant as well. Do I use the angular acceleration I get for (a) and just multiply it by the radius? How do I get angular acceleration?
To solve this problem, we'll use the following equations:
(a) The rotational kinetic energy of the flywheel can be expressed as:
KE = (1/2) I ω^2,
where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
Since the angular velocity is constant, the angular acceleration is zero. Therefore, the rotational kinetic energy is constant.
The equation can be rearranged as:
I ω^2 = constant.
(b) To determine the number of revolutions made before stopping, we need to find the angular displacement.
The angular displacement, θ, can be calculated using the equation:
θ = ω_i t + (1/2) α t^2,
where ω_i is the initial angular velocity, α is the angular acceleration, and t is the time taken.
However, since the angular acceleration is zero during the slowdown, the equation simplifies to:
θ = ω_i t.
We know the initial angular velocity ω_i = 220 rev/min and the time t = 2.7 h.
So, θ = 220 rev/min × 2.7 h = 594 rev.
(c) To find the tangential component of the linear acceleration at a given angular velocity, we can use the equation:
a_t = r α,
where a_t is the tangential acceleration, r is the distance from the axis of rotation, and α is the angular acceleration.
Since the angular acceleration α is zero during the slowdown, this equation simplifies to:
a_t = 0.
Therefore, the tangential component of linear acceleration is zero.
(d) Since the tangential component of linear acceleration is zero, the magnitude of the net linear acceleration will be equal to the centripetal acceleration, given by:
a_c = r ω^2,
where a_c is the centripetal acceleration, r is the distance from the axis of rotation, and ω is the angular velocity.
To find the magnitude of the net linear acceleration, we substitute r = 47 cm (0.47 m) and ω = 110 rev/min into the equation:
a_c = (0.47 m) × (110 rev/min)^2.
Now, let's calculate the solutions:
(a) Since the rotational kinetic energy is constant, the angular acceleration is zero during the slowdown.
(b) The number of revolutions made before stopping is 594 rev.
(c) The tangential component of the linear acceleration is zero.
(d) The magnitude of the net linear acceleration is:
a_c = (0.47 m) × (110 rev/min)^2 = 658 m/min^2