Question

A wheel 1.70 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 3.60 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s. What is the tangential speed, total acceleration, and angular position of point P.

Answer 1

Answer:

Total acceleration will be $30.70m/sec^2$

Explanation:

We have given initial tangential acceleration $\alpha =3.60rad/sec^2$

Radius r = 1.70 m

Initial angular velocity $\omega =0rad/sec$

Time t = 2 sec

From first equation of motion

$\omega _f=\omega +\alpha t$

$\omega _f=0 +3.6\times 2=7.2rad/sec$

Radial acceleration is equal to $a_r=\frac{v^2}{r}=\frac{7.2^2}{1.7}=30.49m/sec^2$

So total acceleration will be equal to

$a=\sqrt{3.6^2+30.49^2}=30.70m/sec^2$