An object whose moment of inertia is 2.5 kg · m^2 experiences the torque shown in the figure below. What is the object's angular velocity at t = 1.6 s? Assume it starts from rest.
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So far, I know that that to find the acceleration, it's T/I (or 2/2.5) and then I just have to multiply the time to get the angular velocity, but I keep getting the wrong answer. What am I doing wrong?
Well, it seems like you're on the right track, but don't worry, I'm here to help you figure out what went wrong. Let's break it down step by step.
First, you correctly determined the torque by dividing 2 N·m by the moment of inertia of 2.5 kg · m^2, which gives you an acceleration of 0.8 rad/s^2.
Now, to find the angular velocity at t = 1.6 s, we need to use the kinematic equation:
angular velocity = initial angular velocity + (acceleration × time)
Since the object starts from rest, the initial angular velocity is 0. Plugging in the values, we get:
angular velocity = 0 + (0.8 rad/s^2 × 1.6 s)
Simplifying, we find that the angular velocity at t = 1.6 s is 1.28 rad/s.
If you got a different answer, maybe there was an error in your calculations, or perhaps a typo. Double-check your math, and hopefully you'll find the correct answer this time!
To find the angular velocity at a given time using the torque and moment of inertia, you need to use the equation:
α = τ / I
where α is the angular acceleration, τ is the torque, and I is the moment of inertia.
In this case, you mentioned that the torque is 2 N·m and the moment of inertia is 2.5 kg·m^2. Thus, we can calculate the angular acceleration as follows:
α = τ / I = 2 N·m / 2.5 kg·m^2 = 0.8 rad/s^2
Given that the object starts from rest, we can further use the equation:
ω = ω0 + α * t
where ω is the final angular velocity, ω0 is the initial angular velocity (which is 0 since it starts from rest), α is the angular acceleration, and t is the time.
Plugging in the values, we can solve for ω:
ω = 0 + (0.8 rad/s^2) * (1.6 s) = 1.28 rad/s
Therefore, the object's angular velocity at t = 1.6 s is 1.28 rad/s.
To find the object's angular velocity at t = 1.6 s, you are correct that you need to calculate the acceleration first and then multiply it by the time.
However, it seems you have made an error in calculating the acceleration. The formula for torque is T = Iα, where T represents the torque, I represents the moment of inertia, and α represents the angular acceleration. Rearranging this equation, we get α = T/I.
In this case, the torque T is given by the figure in the link you provided. However, since I cannot access external links, I cannot view the figure directly. Therefore, I cannot provide you with an accurate value for T.
To calculate the angular acceleration, you need to determine the time rate of change of the torque. This can be done by finding the slope (change in torque divided by the change in time) from the figure depicting the torque. Once you have the angular acceleration, you can multiply it by the given time (t = 1.6 s) to obtain the angular velocity.
Please double-check the torque figure and ensure that you have taken the correct slope to calculate the angular acceleration accurately.
no
your torque is not constant from 0 to 1 second
it is
Torque = 2 t
for 0 to 1 second
a = 2 t/2.5
v = (2/2.5)(t^2/2) = 1/2.5 rad/s at 1 second
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NOW that is your constant torque of 2 from 1 second to 2 seconds
initial v = 1/2.5 rad/s
a = dv/dt = 2/2.5
v = 1/2.5 + (2/2.5)1
v = 3/2.5