Question

A circular saw blade 0.6 m in diameter starts from rest and accelerates with constant acceleration to an angular velocity of 100 rad.s^-1 in 20 s.

Find the angular acceleration.

Determine the angle through which the blade has turned in the 20-s interval.

Solve for the linear velocity of a point in the rim of the saw when the angular velocity is 10 rad.s^-1.

Answer 1

$\vec\omega(t)=\vec\omega(0)+\vec\alpha t;$

$\vec\alpha=\frac{\vec\omega(t)-\vec\omega(0)}{t};$

$\vec\alpha=\frac{\vec\omega(20)-\vec\omega(0)}{20}=\frac{100-0}{20}=5\ rad/s^2$

$\theta_t-\theta_0= \frac{\vec\omega_0+\vec\omega_t}{2}t$

$\Delta\theta= \frac{0+100}{2}*20=1000\ rad$

$\vec v=\vec r\vec\omega;$

$\text{The linear velocity vector is on the tangent to the circle, }\newline \text{the tangent to the circle is orthogonal with the radius}$

$|\vec v|=|\vec r||\vec \omega|\sin90\degree;$

$|\vec v|=10*0.6*1=6m/s$

$\text{Answer: }\vec\alpha=5\ rad/s^2;\ \Delta\theta=1000\ rad;\ |\vec v|=6m/s$