a pulley with rotational inertia of 1.5 times 10 ^-3 kg times m^2 about its axle and a radius of 15 cm is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F=.5t + .3 t^2, where F is in newtons and t in seconds. The pulley is initiall at rest. At t=2s what is the angualr veloctiy?
First we calculate the torque.
t=rxF=r.F.sin(b) where is the angle between r and F (which in this case equals 90°).
So in this case t= r.F = 0,15m.F = 0.075t + 0.045t²
We also know that I = 1.5 . 10^-3 kg.m²
and so we can calculate the angular acceleration (a) with the formula:
t = I.a
=> a = t/I = 50t + 30t²
Now, we need to find the angular velocity after 2 seconds. Since a=dw/dt (with w, the angular velocity), we can find the angular by taking the definite integral of a for t from 0 to 2
=> w = integral of 50t+30t² from 0 to 2=
(25.(2)²+10(2)^3)-(25.(0)²+10(0)^3)= 180-0 = 180
so the angular velocity after two seconds = 180 rad/s
thank you soooooooo much i really appreciate you taking your time to help
To find the angular velocity of the pulley at t = 2s, we need to determine the torque applied to the pulley and use it to calculate the angular acceleration. Then, we can integrate the angular acceleration over time to obtain the angular velocity.
Step 1: Calculate the Torque (τ)
The torque applied to the pulley is given by the equation: τ = I * α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, the moment of inertia is given as 1.5 * 10^-3 kg*m^2.
Step 2: Calculate the Angular Acceleration (α)
To find the angular acceleration, we need to differentiate the force equation with respect to time (t) and divide it by the moment of inertia. Given that the force varies with time according to the equation: F = 0.5t + 0.3t^2, where F is in Newtons and t is in seconds, the derivative of this equation will give us the time-dependent angular acceleration.
Differentiating F = 0.5t + 0.3t^2 with respect to t, we get:
dF/dt = 0.5 + 0.6t
Dividing dF/dt by the moment of inertia (I), we get:
α = (0.5 + 0.6t) / I
Step 3: Find the Angular Velocity (ω)
To find the angular velocity (ω) at t = 2s, we need to integrate the angular acceleration (α) over time from t = 0s to t = 2s.
Integrating α = (0.5 + 0.6t) / I with respect to t, we get:
ω = ∫[(0.5 + 0.6t) / I] dt
= (0.5/I)∫dt + (0.6/I)∫t dt
Evaluating the integrals and plugging in the values:
ω = (0.5/I)∫dt + (0.6/I)∫t dt
= (0.5/I)t + (0.6/I)*(t^2/2) + C
Here, C is the constant of integration. Since the pulley is initially at rest, its initial angular velocity (ω0) is zero. Therefore, we can solve for C by substituting t = 0 and ω = 0 into the equation:
0 = (0.5/I)*0 + (0.6/I)*(0^2/2) + C
C = 0
Plugging C = 0 back into the equation, we get:
ω = (0.5/I)t + (0.6/I)*(t^2/2)
Substituting the given values, I = 1.5 * 10^-3 kg*m^2 and t = 2s:
ω = (0.5/(1.5 * 10^-3))*2 + (0.6/(1.5 * 10^-3))*(2^2/2)
Calculating the expression, we find:
ω = 666.67 rad/s
Therefore, at t = 2s, the angular velocity of the pulley is 666.67 rad/s.