A force FT is applied to a cord wrapped around a pulley with moment of inertia I = 0.435 kg·m2 and radius R0 = 33.0 cm. (See the figure.) There is a frictional torque τfr at the axle of 1.10 m·N. Suppose the force FT is given by the relation FT = 4.00t - 0.10t2 (newtons) where t is in seconds. If the pulley starts from rest, what is the linear speed of a point on its rim 9.0 s later?
a = angular acceleration
I = rotational inertia
w = angular velocity
v = linear velocity
Torque = RxFT = Ia
so,
a = RxFT/I = (R/I)(4t-0.10t^2)
w = integral(a) with respect to t
v= wR
To find the linear speed of a point on the rim of the pulley after 9.0 seconds, we need to calculate the angular acceleration and angular velocity of the pulley.
1. First, we need to find the net torque acting on the pulley. The net torque (τnet) can be calculated as the difference between the applied torque (τapp) and the frictional torque (τfr):
τnet = τapp - τfr
2. The applied torque (τapp) can be calculated using the force (FT) and the radius of the pulley (R0). Since the force FT is given as a function of time, we can substitute the given expression for FT into the equation:
τapp = FT * R0 = (4.00t - 0.10t^2) * 0.33
3. The moment of inertia (I) and the angular acceleration (α) are related by the equation:
τnet = I * α
4. Substituting the expressions for τnet and τapp, and rearranging the equation, we can solve for α:
(4.00t - 0.10t^2) * 0.33 - 1.10 = 0.435 * α
5. Now, we can solve the equation for α by substituting t = 9.0 s and solving for α:
(4.00*9.0 - 0.10*9.0^2) * 0.33 - 1.10 = 0.435 * α
α ≈ 2.036 rad/s^2
6. Next, we can find the angular velocity (ω) at t = 9.0 s by integrating the angular acceleration α with respect to time:
ω = ∫ α dt = ∫ 2.036 dt
ω ≈ 2.036 t + C
7. To find the constant C, we can use the initial condition that the pulley starts from rest (ω = 0 at t = 0):
0 = 2.036 * 0 + C
C = 0
So, the equation for the angular velocity becomes:
ω ≈ 2.036 t
8. Finally, we can find the linear speed (v) of a point on the rim of the pulley by multiplying the angular velocity (ω) by the radius of the pulley:
v = ω * R0 = (2.036 t) * 0.33
v ≈ 0.672 t
Therefore, the linear speed of a point on the rim of the pulley 9.0 seconds later is approximately 0.672 times the time in seconds.
To find the linear speed of a point on the rim of the pulley, we need to calculate the angular speed of the pulley first. Then we can use the formula for linear speed, which relates the angular speed to the radius of the pulley.
Given that the moment of inertia of the pulley is I = 0.435 kg·m^2, the radius of the pulley is R0 = 33.0 cm = 0.33 m, and there is a frictional torque at the axle of τ_fr = 1.10 m·N.
The force applied to the cord wrapped around the pulley, FT, is given by FT = 4.00t - 0.10t^2 (newtons), where t is in seconds.
Since the pulley starts from rest, its initial angular velocity ω_0 is zero. We can write the torque equation for the pulley as:
τ_net = I * α,
where τ_net is the net torque acting on the pulley and α is the angular acceleration.
The net torque acting on the pulley is the difference between the applied torque from the force FT and the frictional torque τ_fr:
τ_net = FT * R0 - τ_fr.
Substituting the expression for FT, we have:
τ_net = (4.00t - 0.10t^2) * R0 - τ_fr.
Now, using the relationship between torque and angular acceleration (τ = I * α), we can write:
(4.00t - 0.10t^2) * R0 - τ_fr = I * α.
Since the angular speed ω is the time derivative of the angular position θ with respect to time (ω = dθ/dt), we can write the angular acceleration as α = dω/dt.
Differentiating the expression for ω with respect to time, we have:
dω/dt = d/dt(dθ/dt) = d^2θ/dt^2.
So, the equation becomes:
(4.00t - 0.10t^2) * R0 - τ_fr = I * (d^2θ/dt^2).
Now, we can solve this differential equation using the initial condition that the pulley starts from rest. Integrating both sides, we get:
(4.00t - 0.10t^2) * R0 - τ_fr = I * ω.
Rearranging the equation, we have:
ω = ((4.00t - 0.10t^2) * R0 - τ_fr) / I.
Now, we can substitute the value of t = 9.0 s into this equation to find the angular speed at that time:
ω = ((4.00 * 9.0 - 0.10 * 9.0^2) * 0.33 - 1.10) / 0.435.
Simplifying this expression, we can calculate the angular speed ω.
Once we have the angular speed ω, we can use the linear speed formula to find the linear speed of a point on the rim of the pulley:
v = ω * R0.
Substituting the values of ω and R0, we can then calculate the linear speed of the point on the pulley's rim 9.0 seconds later.