Radial Component of a wheel?
Obviously the answer for tangential has to be wrong as its unit itself is wrong. The radial acceleration Ar (which changes the direction of motion from the straight line to a circular, tending to the center) is Ar = R*w^2, where R is the radius, w- angular velocity. 130 rpm = (130*2*pi)/60 =13.61357rad/s, which is the initial w. As you have found, the angular acceleration = {[(370-130)/1.1]*2*pi}/60= 6.613879 rad/s^2, so in 1.1s the angular velocity will stand 13.61357+ 6.613879*1.1 = 20.88884rad/s. Placing in the above Ar = 0.66*(20.88884)^2 = 287.9868m/s^2. Also linear acceleration = angular acceleration*R = 6.61*0.66 = 4.36 m/s^2
Determine the radial component of the linear acceleration of a point?
A 64-cm-diameter wheel accelerates uniformly about its center from 110 rpm to 270 rpm rpm in 4.2 s . From a previous part of the problem, I have solved that the angular acceleration (α) is approx. 4.0 rad/s^2. 1) Determine the radial component of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating. 2) Determine the tangential component of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating. All help would be greatly appreciated thanks!
Determine the radial component & tangential component of the linear acceleration?
convert rpm to radians per second (let them equal w1 and w2 respectively): initial theta(dot) = w1 r/s final theta(dot) = w2 r/s Acceleration = (final-initial)/time A) at t=1.9s if acceleration = velocity/time find the velocity of the rotation assuming 'radial' acceleration is the normal acceleration normal acceleration = radius*(angular velocity)^2 B) tangential acceleration = radius*angular acceleration(already calculated) as acceleration is uniform there is no change over time.