A sanding disk with rotational inertia 1.1 x 10^-3 kg*m^2 is attached to an electric drill whose motor delivers a torque of 6 Nm about the central axis of the disk. What are the following values about the central axis at the instant the torque has been applied for 185 milliseconds?
(a) the angular momentum of the disk
_______ kg*m^2/s
(b) the angular speed of the disk
_______ rad/s
--------
my submitted answers were wrong.. and this is what i did/come up with.
185 ms = 0.185 seconds
2pi rad / 0.185 s = 33.96 rad/s (wf)
L = Iwf
L = (1.1 x 10^-3)(33.96)
L = 0.037356
.. am i missing something? .. what am i doing wrong? please show me, thanks!
You are missing the basic laws of motion.
Torque= inerita* angular acceleration
solve for angular acceleration, then.
wf= wi+ angular acceleration*time
angular momentum= I*wf
To solve this problem, you need to use the equations of rotational motion.
First, let's calculate the angular acceleration (α) using the torque equation:
Torque (τ) = Inertia (I) * Angular Acceleration (α)
Given that the torque (τ) is 6 Nm and the inertia (I) is 1.1 x 10^-3 kg*m^2, we can rearrange the equation to solve for α:
α = τ / I
= 6 Nm / (1.1 x 10^-3 kg*m^2)
= 5454.55 rad/s^2
Next, we can use the equation for angular speed (ω):
ω = ω_initial + α * time
Given that the initial angular speed (ω_initial) is 0 rad/s and the time (t) is 0.185 s, we can calculate ω:
ω = 0 + 5454.55 rad/s^2 * 0.185 s
= 1008.18 rad/s
Finally, we can calculate the angular momentum (L) using the equation:
L = I * ω
Given that the inertia (I) is 1.1 x 10^-3 kg*m^2 and the angular speed (ω) is 1008.18 rad/s, we can calculate L:
L = (1.1 x 10^-3 kg*m^2) * 1008.18 rad/s
= 1.109998 kg*m^2/s (rounding to appropriate significant figures)
So, the answers to the questions are:
(a) The angular momentum of the disk is approximately 1.11 kg*m^2/s.
(b) The angular speed of the disk is approximately 1008.18 rad/s.