Question

A turntable rotates with a constant 1.85 rad/s2 clockwise angular acceleration. After 4.00 s it has rotated through a clockwise angle of 30.0 rad . Part A What was the angular velocity of the wheel at the beginning of the 4.00 s interval?

Answer 1

Answer:

wo = 3.34 rad/s

Explanation:

In order to calculate the initial angular velocity of the turntable, you first use the following formula for the angular distance:

$\theta=\omega_o t+\frac{1}{2}\alpha t^2$     (1)

θ: angle = 30.0 rad

wo: initial angular velocity = ?

t: time = 4.00s

α: angular acceleration

You also use the following formula for the angular acceleration:

$\alpha=\frac{\omega-\omega_o}{2\theta}$     (2)

You replace the equation (2) into the equation (1), and you solve the equation for wo:

$\theta=\omega_o t+\frac{1}{2}*\frac{\omega^2-\omega_o^2}{2\theta}t^2$    (3)

You replace the values of the parameters and obtain a quadratic equation:

$30.0=4.0\omega_o+\frac{(1.85)(4.0)^2}{4(30.0)}-\frac{(4.0)^2}{4(30.0)}\omega_o^2\\\\0.133\omega_o^2-4.0\omega_o+29.75$

You use the quadratic formula to obtain the roots:

$\omega_o_{1,2}=\frac{-(-4.0)\pm \sqrt{(-4.0)^2-4(0.133)(29.75)}}{2(0.133)}\\\\\omega_o_{1,2}=\frac{4\pm0.173}{0.266}\\\\\omega_o_{1}=15.53\\\\\omega_o_2=3.34$

The initial angular velocity can be the lower value, that is, 3.34 rad/s

Answer 2

Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s

Explanation: Please see the attachment below