Question

A 3.00 m long rod (I=1/3ML2) with a mass of 10.0 kg is set so that it pivots around its end. The rod is pulled back until it is horizontal and released. A small spherical ball with a mass of 4.00 kg is placed directly under the pivot point of the rod so that the end of the rod strikes the ball when it is vertical.

What is the angular velocity at the bottom before it hits the ball?

Answer 1

Hi there!

Assuming the rod is pulled back 90° (π/2 radians), we can use angular kinematics to solve.

Using the equation:

`\omega_f^2= \omega_i^2 + 2\alpha \theta`

`\omega_f` = final angular velocity (? rad/sec)

`\omega_i` = initial angular velocity (0 rad/sec)

`\alpha` = angular acceleration (rad/sec², must solve for)

`\theta` = angular displacement (π/2 rad)

Recall Newton Second Law's rotational equivalent:

`\Sigma \tau = I\alpha`

τ = Torque (Nm)

I = Moment of inertia (kgm²)
α = angular acceleration (rad/sec²)

Setting the pivot point at the end, and knowing that the force of gravity works at an object's center of mass (1.5m from the rod's end), we can solve for the angular acceleration.

`\tau= F * r\\\\\tau = W * r = 10(9.8) * 1.5 = 147 Nm`

Now, solve for the moment of inertia given I = 1/3ML² at one of its ends:

`I = (1)/(3)(10)(3^2) = 30 kgm^2`

Solve for the angular acceleration:

`147 = 30\alpha\\\\\alpha = 4.9 (rad)/(sec^2)`

Now, we can use the kinematic equation to solve.

`\omega_f^2 = 0 + 2(4.9)((\pi)/(2)})\\\\\omega_f = \sqrt{2(4.9)((\pi)/(2))} = \boxed{3.923 (rad)/(sec)}`