A discus thrower (with arm length of 1.2 m) starts from rest and begins to rotate counterclockwise with an angular acceleration of 1.8 rad/s2.
(a) How long does it take the discus thrower's speed to get to 3.4 rad/s?
(b) How many revolutions does the thrower make to reach the speed of 3.4 rad/s?
revolutions
(c) What is the linear speed of the discus at 3.4 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?
R=1.2 m, ε=1.8 rad/s²
(a) ω=ε•t => t= ω/ε,
(b) 2πN= ε•t ²/2
N= ε•t ²/4π
(c) v= ω•R
(d) a(τ)= ε•R
(e) a(n)=ω²•R
(f) a=sqrt{ a(τ)²+a(n)²}
To solve this problem, we need to use the equations of rotational motion. Here's how we can find the answers to each part of the question:
(a) To find the time it takes for the discus thrower's speed to reach 3.4 rad/s, we can use the following equation:
ω = ω0 + αt
where ω is the final angular velocity (3.4 rad/s), ω0 is the initial angular velocity (0 rad/s since the thrower starts from rest), α is the angular acceleration (1.8 rad/s^2), and t is the time.
Rearranging the equation, we get:
t = (ω - ω0) / α
Plugging in the values:
t = (3.4 - 0) / 1.8
Calculating this gives:
t ≈ 1.89 seconds
So it takes approximately 1.89 seconds for the discus thrower's speed to reach 3.4 rad/s.
(b) To find the number of revolutions the thrower makes to reach a speed of 3.4 rad/s, we can use the equation:
θ = ω0t + (1/2)αt^2
where θ is the angle in radians, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.
Since the thrower starts from rest (ω0 = 0), the equation simplifies to:
θ = (1/2)αt^2
Since the thrower starts from rest, the initial angle is also zero.
Plugging in the values, we have:
θ = (1/2)(1.8)(1.89^2)
Calculating this gives:
θ ≈ 2.69 radians
Since there are 2π radians in one revolution, we can find the number of revolutions by dividing the angle in radians by 2π:
Number of revolutions = θ / (2π)
Plugging in the angle, we have:
Number of revolutions ≈ 2.69 / (2π)
Calculating this gives:
Number of revolutions ≈ 0.43 revolutions
Therefore, the thrower makes approximately 0.43 revolutions to reach a speed of 3.4 rad/s.
(c) To find the linear speed of the discus at 3.4 rad/s, we can use the equation:
v = ω * r
where v is the linear speed, ω is the angular velocity, and r is the arm length of the discus thrower (1.2 m).
Plugging in the values, we have:
v = 3.4 * 1.2
Calculating this gives:
v ≈ 4.08 m/s
Therefore, the linear speed of the discus at 3.4 rad/s is approximately 4.08 m/s.
(d) To find the tangential acceleration of the discus thrower at this point, we can use the equation:
at = α * r
where at is the tangential acceleration, α is the angular acceleration, and r is the arm length of the discus thrower (1.2 m).
Plugging in the values, we have:
at = 1.8 * 1.2
Calculating this gives:
at ≈ 2.16 m/s^2
Therefore, the tangential acceleration of the discus thrower at this point is approximately 2.16 m/s^2.
(e) To find the magnitude of the centripetal acceleration of the discus thrown, we can use the equation:
ac = ω^2 * r
where ac is the centripetal acceleration, ω is the angular velocity, and r is the arm length of the discus thrower (1.2 m).
Plugging in the values, we have:
ac = 3.4^2 * 1.2
Calculating this gives:
ac ≈ 13.824 m/s^2
Therefore, the magnitude of the centripetal acceleration of the discus thrown is approximately 13.824 m/s^2.
(f) The total acceleration of an object moving in a circular path is the vector sum of the tangential and the centripetal accelerations. Since they are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the total acceleration:
atotal = √(at^2 + ac^2)
Plugging in the values, we have:
atotal = √(2.16^2 + 13.824^2)
Calculating this gives:
atotal ≈ 14.014 m/s^2
Therefore, the magnitude of the discus's total acceleration is approximately 14.014 m/s^2.