The flywheel of a steam engine runs with a constant angular speed of 350 rev/min. When steam is shut off, the friction of the bearings stops the wheel in 1.0 h.
A) At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 50 cm from the axis of rotation?
B) What is the magnitude of the net linear acceleration of the particle in A?
I calculated the acceleration to be -5.83 rev/s and that it rotates 10500 times before stopping, but I have no idea how to do these two.
To solve these problems, we can start by finding the linear acceleration of the flywheel particle at the instant it is turning at 75 rev/min.
A) To find the tangential component of the linear acceleration, we can use the formula:
a_t = r * α
where a_t is the tangential component of the linear acceleration, r is the distance from the axis of rotation (50 cm in this case), and α is the angular acceleration.
Since the problem states that the angular speed is constant at 350 rev/min, we can convert this to radians per second:
ω = 350 rev/min * (2π rad/rev) * (1 min/60 s)
= 350 * 2π/60 rad/s
= 11.55 rad/s
Now, we can calculate the angular acceleration α by dividing the change in angular speed by the change in time:
α = (ω_f - ω_i) / t
where ω_f is the final angular speed (75 rev/min converted to rad/s) and ω_i is the initial angular speed (350 rev/min converted to rad/s).
ω_i = 350 rev/min * (2π rad/rev) * (1 min/60 s)
= 350 * 2π/60 rad/s
= 11.55 rad/s
ω_f = 75 rev/min * (2π rad/rev) * (1 min/60 s)
= 75 * 2π/60 rad/s
= 2.48 rad/s
Substituting the values into the formula:
α = (2.48 rad/s - 11.55 rad/s) / (1 h converted to s)
= (-9.07 rad/s) / 3600 s
= -0.00252 rad/s^2
Now, plug the values into the formula for tangential acceleration:
a_t = r * α
= 0.50 m * (-0.00252 rad/s^2)
= -0.00126 m/s^2
Therefore, the tangential component of the linear acceleration of the flywheel particle is -0.00126 m/s^2.
B) To find the magnitude of the net linear acceleration of the particle, we can use the formula:
a_net = √(a_t^2 + a_c^2)
where a_net is the net linear acceleration, a_t is the tangential component of the linear acceleration, and a_c is the centripetal component of the linear acceleration.
The centripetal component of the linear acceleration can be calculated using the formula:
a_c = r * ω^2
where r is the distance from the axis of rotation (50 cm in this case) and ω is the angular speed.
Let's calculate the centripetal component of the linear acceleration:
a_c = (0.50 m) * (11.55 rad/s)^2
= 6.50 m/s^2
Now, we can find the net linear acceleration:
a_net = √((-0.00126 m/s^2)^2 + (6.50 m/s^2)^2)
= √(1.5876 x 10^-6 m^2/s^4 + 42.25 m^2/s^4)
= √42.2500015876 m^2/s^4
= 6.50 m/s^2
Therefore, the magnitude of the net linear acceleration of the particle is 6.50 m/s^2.
To answer these questions, we need to use some concepts from rotational motion and kinematics.
A) To find the tangential component of linear acceleration, we can use the formula for linear acceleration in circular motion:
a_tangential = r * α
where:
- a_tangential is the tangential component of linear acceleration
- r is the distance from the axis of rotation (in this case, 50 cm)
- α is the angular acceleration
Since the problem states that the flywheel runs with a constant angular speed, we know that the angular acceleration is zero. Therefore, the tangential component of linear acceleration is also zero.
B) To find the magnitude of the net linear acceleration of the particle, we need to use the formula for linear acceleration in circular motion:
a_net = r * ω^2
where:
- a_net is the net linear acceleration
- r is the distance from the axis of rotation (in this case, 50 cm)
- ω is the angular speed
Given that the angular speed (ω) is 75 rev/min, we first need to convert it to rad/s. Since 1 revolution is equal to 2π radians, we have:
ω = (75 rev/min) * (2π rad/rev) * (1 min/60 s)
Now we can calculate the net linear acceleration:
a_net = (50 cm) * [(75 rev/min) * (2π/rev) * (1/60 s)]^2
Make sure to convert the units appropriately so that the final answer is in the desired units.
Let's plug in the values and calculate the answer:
a_net = (50 cm) * [(75 rev/min) * (2π/rev) * (1/60 s)]^2
= (50 cm) * [(75 * 2π/60)^2 cm/s^2]
= (50 cm) * [(75 * 2π/60)^2 cm/s^2]
= (50 cm) * [(75 * 2 * 3.14/60)^2 cm/s^2]
= (50 cm) * [(2 * 15.7/60)^2 cm/s^2]
= (50 cm) * [(0.523)^2 cm/s^2]
= (50 cm) * [0.273cm/s^2]
≈ 13.65 cm/s^2
Therefore, the magnitude of the net linear acceleration of the particle (at a distance of 50 cm from the axis of rotation) is approximately 13.65 cm/s^2.