A disk with a rotational inertia of 7 kgm^2 rotates like a merry-go-round while undergoing a torque given by (5+2t)Nm. At time 1s, its angular momentum is 5kg m^2/s. what is its angular momentum at 3s?

I can't figure out how to start this problem.

angular momentum change is the integral of torque, which is a time-dependent functiuon in this case.

(I*w)@t=3s - (I*w)@t=1s
= Integral (5 + 2t) dt
t = 1 to t = 3
= 5*(3-1) + (3^2 - 1^2)
= 10 + 8 = 18
Add 5 (the initial angualr momentum) to that for the angular momentum at 3 s.

Thank you

To solve this problem, we can use the definition of torque and angular momentum.

The torque acting on the disk is given by (5+2t) Nm, where t is the time in seconds.

The formula for torque is the product of the force acting on an object and the distance from the axis of rotation. In this case, the force is not given, but we can assume it is acting at a distance of 1 meter from the axis of rotation.

Therefore, torque = force × distance.
(5+2t) Nm = force × 1m

To find the angular momentum at a given time, we can use the equation:
Angular momentum = rotational inertia × angular velocity

Given that the rotational inertia of the disk is 7 kgm^2, and at time 1s, the angular momentum is 5 kgm^2/s, we can use this information to solve for the angular velocity.

5 kgm^2/s = 7 kgm^2 × angular velocity
angular velocity = (5 kgm^2/s) / (7 kgm^2)

Now we have the angular velocity at time 1s. To find the angular momentum at 3s, we can use the equation:
Angular momentum = rotational inertia × angular velocity

Let's calculate the angular momentum at 3s:

angular velocity at 3s = angular velocity at 1s + (change in angular velocity per second × time)
angular velocity at 3s = (5 kgm^2/s) / (7 kgm^2) + (2 kgm^2/s^2 × 2s)
angular velocity at 3s = 5/7 + 4/7 = 9/7 kgm^2/s

angular momentum at 3s = rotational inertia × angular velocity
angular momentum at 3s = 7 kgm^2 × (9/7 kgm^2/s)
angular momentum at 3s ≈ 63/7 kgm^2/s
angular momentum at 3s ≈ 9 kgm^2/s

Therefore, the angular momentum of the disk at 3s is approximately 9 kgm^2/s.

To approach this problem, you need to use the concept of torque and angular momentum. Here's how you can solve it step by step:

Step 1: Find the angular acceleration (α):
The torque (τ) is given by (5+2t) Nm, where t is time in seconds. Torque is related to angular acceleration (α) and inertia (I) by the equation τ = I * α.

Since the torque is given as (5+2t) Nm and the rotational inertia (I) is given as 7 kgm^2, substitute these values into the equation: (5+2t) = 7 * α.

Step 2: Solve for angular acceleration (α):
Rearrange the equation to solve for α: α = (5+2t)/7.

Step 3: Find the initial angular velocity (ω1):
The angular momentum (L) of the disk at time t is given as 5 kgm^2/s. Angular momentum is equal to the product of rotational inertia (I) and angular velocity (ω).

So, L = I * ω1, where ω1 is the initial angular velocity.

Rearrange the equation to solve for ω1: ω1 = L/I.

Substitute the given values of L = 5 kgm^2/s and I = 7 kgm^2 into the equation: ω1 = 5/7.

Step 4: Find the angular velocity (ω3) at time t = 3s:
To find ω3 at time t = 3s, you need to calculate the change in angular velocity (∆ω) and add it to the initial angular velocity (ω1).

The change in angular velocity (∆ω) can be obtained using the equation ∆ω = α * ∆t, where α is the angular acceleration and ∆t is the change in time. Here, ∆t = 3s - 1s = 2s.

Substitute the values of α = (5+2t)/7 and ∆t = 2s into the equation: ∆ω = [(5+2t)/7] * 2.

Now, add the change in angular velocity (∆ω) to the initial angular velocity (ω1) to find ω3: ω3 = ω1 + ∆ω.

Substitute the values of ω1 = 5/7 and ∆ω = [(5+2t)/7] * 2 into the equation.

Finally, simplify the expression to find the angular momentum at t = 3s.

I hope that helps!