A discus thrower (with arm length of 1.2 m) starts from rest and begins to rotate counterclockwise with an angular acceleration of 3.0 rad/s2.
(a) How long does it take the discus thrower's speed to get to 4.5 rad/s?
(b) How many revolutions does the thrower make to reach the speed of 4.5 rad/s?
revolutions
(c) What is the linear speed of the discus at 4.5 rad/s?
(d) What is the tangential acceleration of the discus thrower at this point?
(e) What is the magnitude of the centripetal acceleration of the discus thrown?
(f) What is the magnitude of the discus's total acceleration?
I will be happy to critique your thinking on this.
a) 4.5/3= 1.5 s
b) i'm still confused of how to get the number of revolutions
c) a=v^2/r the answer is 5.4 m/s
d) at=r alpha but we need to find alpha first to get at
b, displacement= 1/2 angacceleration*time^2
c. Your answer is nonsense. Tangential velocty = angularvelocity*r
d.total acceleartion: find centripetal acceleration (v^2/r) That is inward. No tangentail acceleration is 3*r
those are at ninety degrees, add them as vectors (Pythagorean theorem_
i just want to find the number of revolutions , i know that 2 pi rad= 1 rev but how do we find that in this question.
To answer these questions, we need to understand the relationship between angular and linear motion.
(a) To find the time it takes for the thrower's speed to reach 4.5 rad/s, we can use the formula for angular acceleration:
angular acceleration (α) = change in angular velocity (Δω) / time (t)
Rearranging the formula, we have:
Δω = α * t
Given that the angular acceleration (α) is 3.0 rad/s^2 and we want to find the time it takes to reach a final angular velocity (ω) of 4.5 rad/s, we can substitute the values:
4.5 rad/s = 3.0 rad/s^2 * t
Simplifying the equation, we find:
t = 4.5 rad/s / 3.0 rad/s^2 = 1.5 s
Therefore, it takes 1.5 seconds for the thrower's speed to reach 4.5 rad/s.
(b) To find the number of revolutions the thrower makes to reach a speed of 4.5 rad/s, we can use the formula for angular displacement:
angular displacement (θ) = initial angular velocity (ω_0) * time (t) + 1/2 * angular acceleration (α) * time squared (t^2)
In this case, the thrower's initial angular velocity (ω_0) is 0 rad/s since they start from rest. And we want to find the time it takes to reach a final angular velocity (ω) of 4.5 rad/s, which we calculated to be 1.5 seconds in part (a).
Substituting the values into the formula:
θ = 0 rad/s * 1.5 s + 1/2 * 3.0 rad/s^2 * (1.5 s)^2
Simplifying the equation, we find:
θ = 1.6875 rad
Since there are 2π radians in one revolution, we can find the number of revolutions by dividing the angular displacement by 2π:
Number of revolutions = 1.6875 rad / (2π rad/rev) ≈ 0.268 rev
Therefore, the thrower makes approximately 0.268 revolutions to reach a speed of 4.5 rad/s.
(c) To find the linear speed of the discus at 4.5 rad/s, we can use the formula:
linear speed (v) = radius (r) * angular velocity (ω)
Given that the radius of the discus thrower's arm is 1.2 m, we can substitute the values:
v = 1.2 m * 4.5 rad/s ≈ 5.4 m/s
Therefore, the linear speed of the discus at 4.5 rad/s is approximately 5.4 m/s.
(d) The tangential acceleration (at) can be found using the formula:
tangential acceleration (at) = radius (r) * angular acceleration (α)
Substituting the given values:
at = 1.2 m * 3.0 rad/s^2 = 3.6 m/s^2
Therefore, the tangential acceleration of the discus thrower at this point is 3.6 m/s^2.
(e) The centripetal acceleration (ac) can be found using the formula:
centripetal acceleration (ac) = radius (r) * (angular velocity (ω))^2
Substituting the given values:
ac = 1.2 m * (4.5 rad/s)^2 ≈ 24.3 m/s^2
Therefore, the magnitude of the centripetal acceleration of the discus thrower is approximately 24.3 m/s^2.
(f) The total acceleration (a) can be found using the formula:
total acceleration (a) = √(tangential acceleration (at))^2 + (centripetal acceleration (ac))^2
Substituting the given values:
a = √(3.6 m/s^2)^2 + (24.3 m/s^2)^2 ≈ 24.6 m/s^2
Therefore, the magnitude of the discus's total acceleration is approximately 24.6 m/s^2.