please help...A wheel 1.0m in diameter is rotating about a fixed axis with an initial angular velocity of 2rev.s-1.the angular acceleration is 3rev.s-1.
(1)compute the angular velocity after 6s.
(2)through what angle has the wheel turned in this time interval?
(3)what is the tangenial velocity of a point on the rim of the wheel at t=6s?
The units of angular velocity have to be rad/s and the units of angular acceleration have to be rad/s^2.
For (1), you can convert rev/s to rad/s, but your units of angular acceleration are entirely wrong. Time has to be to the -2 power.
Review the problem statement and make sure you got it right.
im sorry that was a typing error for acceleration...i have converted rev/s to rad/s in question 1..i am not getting the correct answers ,im not sure if it is the equations i am using ,could you please help me figure these questions out..thank you
really appreciate your help
answer please
Sure, I'd be happy to help you with these questions!
In order to answer these questions, we need to use some basic equations related to rotational motion. We'll use the following equations:
1. Angular velocity (ω) = initial angular velocity + angular acceleration * time (ω = ω0 + αt)
2. Angle (θ) = initial angular velocity * time + 0.5 * angular acceleration * time^2 (θ = ω0t + 0.5αt^2)
3. Tangential velocity (v) = radius * angular velocity (v = rω)
Let's calculate the answers step by step:
(1) Compute the angular velocity after 6s:
Using the formula ω = ω0 + αt, where ω0 = 2 rev/s and α = 3 rev/s^2, and t = 6s:
ω = 2 rev/s + 3 rev/s^2 * 6s
= 2 rev/s + 18 rev/s
= 20 rev/s
Therefore, the angular velocity after 6s is 20 rev/s.
(2) Through what angle has the wheel turned in this time interval?
Using the formula θ = ω0t + 0.5αt^2, where ω0 = 2 rev/s, α = 3 rev/s^2, and t = 6s:
θ = 2 rev/s * 6s + 0.5 * 3 rev/s^2 * (6s)^2
= 12 rev + 0.5 * 3 rev/s^2 * 36 s^2
= 12 rev + 54 rev
= 66 rev
Therefore, the wheel has turned 66 revolutions in this time interval.
(3) What is the tangential velocity of a point on the rim of the wheel at t = 6s?
Using the formula v = rω, where r is the radius of the wheel and ω is the angular velocity. We know that the diameter of the wheel is 1.0m, so the radius (r) is 0.5m. Plugging in the values:
v = 0.5m * 20 rev/s
= 10 m/s
Therefore, the tangential velocity of a point on the rim of the wheel at t = 6s is 10 m/s.
I hope this helps! Let me know if you have any further questions.