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)?
to get angular deceleration, take the angular velocity and divide by the time. Watch units: I recommend change hrs, min to seconds, and revs to radians.
To answer these questions, we will need to use the concepts of angular velocity, angular acceleration, and linear acceleration. Let's solve them one by one:
(a) To find the constant angular acceleration, we can use the formula:
Angular acceleration (α) = Change in angular velocity (ω) / Time taken (t)
Here, the change in angular velocity is from 220 rev/min to 0 rev/min, and the time taken is given as 2.7 hours. However, we need to convert the time into minutes:
2.7 hours × 60 minutes/hour = 162 minutes
Now, we can calculate the angular acceleration:
α = (0 - 220 rev/min) / 162 min
Simplifying the equation:
α = -1.358 rev/min²
The negative sign indicates that the angular acceleration is in the opposite direction of the initial angular velocity.
(b) To find the number of revolutions the wheel makes before stopping, we can use the formula for angular displacement:
Angular displacement (θ) = Initial angular velocity (ω₀) × Time taken (t) + 0.5 × Angular acceleration (α) × (Time taken (t))²
Since the final angular velocity is zero, we can rearrange the formula as:
θ = ω₀t + 0.5αt²
Substituting the values:
θ = (220 rev/min) × (162 min) + 0.5 × (-1.358 rev/min²) × (162 min)²
Simplifying the equation:
θ = 32599.178 rev
Therefore, the wheel makes approximately 32599 revolutions before stopping.
(c) To find the tangential component of the linear acceleration at a given instant, we need to use the formula:
Tangential linear acceleration (a_t) = Radius (r) × Angular acceleration (α)
Substituting the given values:
a_t = (47 cm) × (-1.358 rev/min²)
To convert the angular acceleration to rad/min², we multiply it by 2π:
a_t = (47 cm) × (-1.358 rev/min²) × (2π rad/rev) × (1 min/60 s)
Simplifying the equation:
a_t ≈ -2.262 cm/s²
Therefore, the tangential component of the linear acceleration of the flywheel particle is approximately -2.262 cm/s².
(d) To find the magnitude of the net linear acceleration, we need to consider both the tangential and centripetal components. The centripetal acceleration can be calculated using the formula:
Centripetal linear acceleration (a_c) = Radius (r) × Angular velocity² (ω²)
Substituting the given values:
a_c = (47 cm) × (110 rev/min)² × (2π rad/rev)² × (1 min/60 s)²
Simplifying the equation:
a_c ≈ 98.413 cm/s²
To find the magnitude of the net linear acceleration, we can use the Pythagorean theorem:
Net linear acceleration (a) = √(Tangential linear acceleration (a_t)² + Centripetal linear acceleration (a_c)²)
Substituting the values:
a = √((-2.262 cm/s²)² + (98.413 cm/s²)²)
Simplifying the equation:
a ≈ 98.42 cm/s²
Therefore, the magnitude of the net linear acceleration of the particle is approximately 98.42 cm/s².