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?
To solve these questions, we can use the following formulas:
(1) Angular velocity (ω) after time (t) can be calculated using the formula:
ω = ω₀ + αt,
where ω₀ is the initial angular velocity,
α is the angular acceleration, and
t is the time elapsed.
(2) The angle (θ) through which the wheel has turned in time (t) can be calculated using the formula:
θ = ω₀t + (1/2)αt².
(3) The tangential velocity (v) of a point on the rim of the wheel can be calculated using the formula:
v = rω,
where r is the radius of the wheel and
ω is the angular velocity.
Given:
- Diameter (d) of the wheel = 1.0 m,
- Initial angular velocity (ω₀) = 2 rev/s,
- Angular acceleration (α) = 3 rev/s², and
- Time (t) = 6 s.
Let's solve each question:
(1) Angular velocity after 6s:
Using ω = ω₀ + αt,
ω = 2 rev/s + (3 rev/s²)(6 s)
ω = 2 rev/s + 18 rev/s
ω = 20 rev/s.
(2) Angle turned in 6s:
Using θ = ω₀t + (1/2)αt²,
θ = (2 rev/s)(6 s) + (1/2)(3 rev/s²)(6 s)²
θ = 12 rev + 54 rev
θ = 66 rev.
(3) Tangential velocity at t = 6s:
The radius (r) of the wheel can be calculated using r = (d/2).
r = 1.0 m / 2
r = 0.5 m.
Using v = rω,
v = (0.5 m)(20 rev/s)
v = 10 m/s.
Therefore, the answers are:
(1) The angular velocity after 6s is 20 rev/s.
(2) The wheel has turned through an angle of 66 revolutions in this time interval.
(3) The tangential velocity of a point on the rim of the wheel at t = 6s is 10 m/s.