A 51-cm-diameter wheel accelerates uniformly about its center from 150 rpm to 290 rpm in 4.0 s.(A) Determine it's angular acceleration. (B) Determine the radial component of the linear acceleration of a point on the edge of the wheel 1.1 s after it has started accelerating. (C) Determine the tangential component of the linear acceleration of a point on the edge of a wheel 1.1 s after it has started accelerating.

Answer: (A) 7/6 pi /s^2 (B) 4.145 m/s^2(C) 119pi/200 m/s^2Explanation: Part A. The angular acceleration is given by [tex]\alpha=\frac{\omega_f-\omega_i}{\Delta t}[/tex]where wf is the initial angular velocity and wi is the final angular velocity and t is the time interval.Now, we are given the angular velocity is given in rpm and we have to convert it into radians/sec . 150 rpm = 150 x 2pi / 60 min = 5 pi rad/ sec 290 rpm = 290 x 2pi / 60 min = 29/ 3 pi rad/ secNow we are in the position to find the angular acceleration [tex]\alpha=\frac{\frac{29}{3}\pi-5\pi}{4s-0s}[/tex][tex]\boxed{\alpha=\frac{7}{6}\pi\; /s^2}[/tex]which is our answer! Part B. The radial acceleration is given by [tex]a_r=\frac{v^2}{R}[/tex]where v is the velocity of the object (moving in a circle) and R is the radius of the circle. Now, [tex]v=\alpha Rt[/tex]putting in the values of alpha, R and t = 1.1 s gives [tex]v=\frac{7}{6}\pi\times\frac{0.51}{2}\times1.1[/tex][tex]v=1.028\; m/s[/tex]therefore, [tex]a_r=\frac{v^2}{R}=\frac{(1.028)^2}{0.51/2}[/tex][tex]\boxed{a_r=4.145/s^2}[/tex]which is our answer! Part C.Here we have to relationship between angular and tangential acceleration: [tex]a=\alpha R[/tex]where r is the radius of the circle. Since R = 0.51/2 m, we have [tex]a=\frac{7}{6}\pi\cdot\frac{0.51}{2}m[/tex][tex]\boxed{a=\frac{119}{400}\pi}[/tex]which is our answer! Hence, to summerise (A) 7/6 pi /s^2 (B) 4.145 m/s^2(C) 119pi/200 m/s^2