I have two questions, both are about angular momentum so I thought I'd post them together. I'm really confused on anything regarding rolling wheels and things like that (help me please!. It seems as though the whole chapter was amassed of all of the topics I cant wrap my head around. For both of these, the only work I have down is angular momentum l=rXp, or m(rXv), or =rmvsintheta. Any help would be so great and much appreciated. Thanks so much in advance!
1. A particle is acted on by two torques about the origin: τ1 has a magnitude of 4.5 N·m and is directed in the positive direction of the x axis, and τ2 has a magnitude of 1.4 N·m and is directed in the negative direction of the y axis. What are the magnitude and direction of d/dt, where is the angular momentum of the particle about the origin?
2. A sanding disk with rotational inertia 1.3 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of 9 N·m 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 140 ms?
(a) the angular momentum of the disk
in kg·m2/s
(b) the angular speed of the disk
in rad/s
You failed to mention moment of inertia. I recommend you reread that section, it is what imposes resistance to change of angular motion.
Torque=momentinertia*alpha (alpha is angular acceleartion, in radians/sec^2)
Torque*time=momentinertia*change angular velocity
angular speed final=initialangular speed +alpha*time
angular displacement=intial angular speed*time+1/2 alpha *Time^2
MEMORIZE these, understand them. Make flash cards, these are the equations of motion.
2.a)angular momentum =momentinertia (intial w+alpha*time)
but alpha=torque/moment of inertia (the first equation you should memorize, above).
solve.
b wfinal=alpha*time
Sure! I'll be happy to help you with both of these questions.
1. To find the magnitude and direction of d/dt (the time derivative) of the angular momentum of the particle about the origin, we can use the equation:
d/dt L = τ_total
Here, τ_total is the total torque acting on the particle. To find τ_total, we need to add the individual torques, taking into account their directions. Since torque is a vector quantity, we need to consider both magnitude and direction.
Given:
τ1 = 4.5 N·m (positive x-axis direction)
τ2 = -1.4 N·m (negative y-axis direction)
To find τ_total, we can sum the individual torques:
τ_total = τ1 + τ2
Taking into account their directions, we can write:
τ_total = (4.5 N·m) i - (1.4 N·m) j
Now, since we know that d/dt L = τ_total, we can find the magnitude and direction of d/dt L by using the components of τ_total:
|d/dt L| = sqrt[(4.5 N·m)^2 + (-1.4 N·m)^2]
= sqrt[20.25 N^2·m^2 + 1.96 N^2·m^2]
= sqrt[22.21 N^2·m^2]
≈ 4.71 N·m
The direction of d/dt L can be found by finding the angle θ it makes with the x-axis. We can use the arctan function to find this angle:
θ = arctan(-(1.4 N·m) / (4.5 N·m))
Using a calculator, we find:
θ ≈ -17.69 degrees (counterclockwise from the positive x-axis)
Therefore, the magnitude and direction of d/dt L are approximately 4.71 N·m and -17.69 degrees counterclockwise from the positive x-axis, respectively.
2. To find the values about the central axis at the instant the torque has been applied for 140 ms, we can use the equations:
(a) L = Iω
For part (a), we need to find the angular momentum of the disk.
Given:
I = 1.3 * 10^-3 kg·m^2 (rotational inertia)
τ = 9 N·m (torque)
t = 140 ms = 0.14 s (time)
Using the equation L = Iω, we can rearrange it to solve for L:
L = I * (dω/dt)
Since the torque τ = dL/dt, we can substitute it in the equation:
τ = I * (dω/dt)
dω/dt = τ/I
Plugging in the given values, we get:
dω/dt = (9 N·m) / (1.3 * 10^-3 kg·m^2)
≈ 6923 rad/s^2
Now, to find the angular momentum L, we can integrate dω/dt:
L = ∫(dω/dt) dt
L = ∫(6923 rad/s^2) (0 to 0.14 s) (limits of integration)
L = (6923 rad/s^2) * (0.14 s)
≈ 970 kg·m^2/s
Therefore, the angular momentum of the disk about the central axis at the instant the torque has been applied for 140 ms is approximately 970 kg·m^2/s.
(b) To find the angular speed of the disk at the instant the torque has been applied for 140 ms, we can use the equation:
ω = ω_0 + (dω/dt) * t
Given:
ω_0 = 0 rad/s (initial angular speed)
(dω/dt) = 6923 rad/s^2 (rate of change of angular speed)
t = 0.14 s (time)
Plugging in the values, we can find:
ω = 0 rad/s + (6923 rad/s^2) * (0.14 s)
≈ 969 rad/s
Therefore, the angular speed of the disk about the central axis at the instant the torque has been applied for 140 ms is approximately 969 rad/s.
I hope this helps! Let me know if you have any further questions.