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?
Plss naww
Plss naww I need help
acceleration is in s^-2
(1) 2rps + (3 rps^2 * 6 s) = 20 rps
(2) ave rps = (2 + 20) / 2 = 11rps
... 11 rps * 6 s = 66 rev ... 2π rad per rev
(3) tan vel is ... 20 rps * circumference
... circumference is π times diameter
To answer these questions, we'll need to use a few formulas related to rotational motion. Let's start with the given information:
Diameter of the wheel (d) = 1.0 m
Initial angular velocity (ω_i) = 2 rev/s
Angular acceleration (α) = 3 rev/s^2
Time interval (t) = 6 s
Now let's answer the questions one by one:
(1) Compute the angular velocity after 6s:
Angular velocity (ω) is given by the equation:
ω = ω_i + αt
Substituting the values:
ω = 2 rev/s + (3 rev/s^2)(6 s)
= 2 rev/s + 18 rev/s
= 20 rev/s
Therefore, the angular velocity after 6 seconds is 20 rev/s.
(2) Through what angle has the wheel turned in this time interval:
The angle (θ) turned by the wheel is given by the equation:
θ = ω_i * t + 0.5 * α * t^2
Substituting the values:
θ = (2 rev/s)(6 s) + 0.5 * (3 rev/s^2)(6 s)^2
= 12 rev + 0.5 * 3 rev/s^2 * 36 s^2
= 12 rev + 54 rev
= 66 rev
Therefore, the wheel has turned through an angle of 66 revolutions in this time interval.
(3) What is the tangential velocity of a point on the rim of the wheel at t = 6s:
The tangential velocity (v) of a point on the rim of the wheel is given by the equation:
v = ω * r
where r is the radius of the wheel, which is half of the diameter.
Substituting the values:
r = 1.0 m / 2
= 0.5 m
v = (20 rev/s) * (0.5 m/rev)
= 10 m/s
Therefore, the tangential velocity of a point on the rim of the wheel at t=6s is 10 m/s.