A discus thrower accelerates a discus from rest to a speed of 25m/s by whirling it at 1.25 rev. The discus moves in circle of radius 1m. Find a. The final angular speed, b. The constant angular acceleration, c. The time interval from rest to the speed 25m/s.
Thanks for the help!
a) omega = v/r
b) alpha = omega^2/2(theta )
where theta = 1.25*2pi
c) t = omega/alpha (since initial is zero)
To solve this problem, we can use the kinematic equation for rotational motion:
ω_f = ω_0 + αΔt
where:
- ω_f is the final angular velocity
- ω_0 is the initial angular velocity (0, since the discus starts from rest)
- α is the angular acceleration
- Δt is the time interval
Given:
- initial angular velocity (ω_0) = 0
- final angular velocity (ω_f) = 25 m/s
- number of revolutions (n) = 1.25 rev
- radius of the circle (r) = 1 m
a. Final Angular Speed (ω_f):
To convert the given number of revolutions to radians, we can use the conversion factor (2π radians = 1 revolution). The final angular displacement (θ) can be calculated as:
θ = n * 2π
Substituting the given values:
θ = 1.25 * 2π
Now, we can use the formula for linear velocity to relate it to angular velocity:
v = ω * r
where:
- v is the linear velocity
- ω is the angular velocity
- r is the radius of the circle
Substituting the given values:
v = ω_f * r
25 m/s = ω_f * 1
Therefore, the final angular speed (ω_f) is 25 rad/s.
b. Constant Angular Acceleration (α):
To find the constant angular acceleration, we can use the equation:
α = (ω_f - ω_0) / Δt
Since the initial angular velocity (ω_0) is 0:
α = ω_f / Δt
Substituting the given values:
α = 25 rad/s / Δt
c. Time Interval (Δt):
To find the time interval from rest to the speed of 25 m/s, we can rearrange the formula for α:
25 rad/s = α * Δt
Therefore, the time interval (Δt) is:
Δt = 25 rad/s / α
Substituting the previously found value for α:
Δt = 25 rad/s / (25 rad/s / Δt)
Δt = Δt
This means that Δt can have any value since it cancels out. Therefore, no specific time interval can be determined in this case.
To find the final angular speed, constant angular acceleration, and the time interval from rest to a speed of 25m/s, we can use the formulas of circular motion.
a. Final angular speed (ωf):
The final angular speed is calculated using the formula:
ωf = (2π × the number of revolutions) / time taken
Given:
The number of revolutions = 1.25 rev
We need to find the time taken.
To find the time taken, we'll use the formula for linear speed:
v = ω × r
where v is the linear speed, ω is the angular speed, and r is the radius.
In this case, the linear speed v is given as 25m/s, and the radius r is given as 1m. Therefore, we can rearrange the formula to find ω:
ω = v / r
Now we have the angular speed ω, and we can find the time taken using the formula mentioned above:
time taken = (2π × the number of revolutions) / ω
Substituting the given values, we get:
time taken = (2π × 1.25 rev) / (25m/s / 1m) = 0.125π s = 0.393 s
Therefore, the final angular speed (ωf) is not yet calculated. But the time taken from rest to a speed of 25m/s is found to be 0.393 seconds.
b. Constant angular acceleration (α):
Given that the discus accelerates from rest to a final speed, the acceleration is constant.
We can use the formula to calculate angular acceleration (α):
α = (ωf - ω0) / t
where ωf is the final angular speed, ω0 is the initial angular speed (which is 0 in this case since the discus starts from rest), and t is the time interval.
Substituting the values, we get:
α = (ωf - 0) / 0.393 s = ωf / 0.393 s
Therefore, the constant angular acceleration (α) is ωf / 0.393 s.
c. Time interval from rest to a speed of 25m/s:
We have already calculated this value as 0.393 seconds in the previous calculations.
To summarize:
a. The final angular speed (ωf) has not been calculated yet.
b. The constant angular acceleration (α) is ωf / 0.393 s.
c. The time interval from rest to the speed of 25m/s is 0.393 seconds.