a wheel revolves
about a fixed horizontal axis through O (the center).
During a certain interval of time, the
angular velocity (which initially is 20 rad
per second clockwise) changes uniformly
at such a rate that the angular
displacement is 60 rad counterclockwise,
and the total angle turned through is 100
rad. Point A is in the position shown 2.5
sec after the beginning of the time
interval.
Determine, for this instant, the linear
acceleration of point A.
To determine the linear acceleration of point A, we need to first find the angular acceleration of the wheel using the given information.
Angular displacement (θ) = 60 rad (counterclockwise)
Total angle turned through (Δθ) = 100 rad
Initial angular velocity (ω0) = 20 rad/s (clockwise)
Time interval (Δt) = 2.5 sec
We can use the following kinematic equation for rotational motion to find the angular acceleration (α):
θ = ω0*t + 0.5*α*t^2
Substituting the given values:
60 rad = 20 rad/s * 2.5 s + 0.5 * α * (2.5 s)^2
Simplify and solve for α:
60 rad - 50 rad = 0.625 α
10 rad = 0.625 α
α = 10 rad / 0.625
α = 16 rad/s^2
Now that we have the angular acceleration, we can find the linear acceleration of point A using the relation:
a = α * r
where "a" is the linear acceleration, "α" is the angular acceleration, and "r" is the distance of point A from the axis of rotation.
Since we have no information about the distance of point A from the axis of rotation, we cannot determine the exact value of the linear acceleration for this instant. However, we can compute an expression for the linear acceleration in terms of the distance "r".
Please provide the distance of point A from the axis of rotation, and I can help you calculate the linear acceleration for that particular instant.