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Sagot :
Answer:
(A) 7/6 pi /s^2
(B) 4.145 m/s^2
(C) 119pi/200 m/s^2
Explanation:
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/ sec
Now 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
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