Question

Question: A disk with a rotational inertia of 7.00 kg m²rotates like a merry-go-round while undergoing a variable torque given$\mathit{b}\mathit{y}\mathit{\tau }\mathbf{=}\left(5.00+2.00t\right)\mathit{N}\mathit{m}$. At time, t = 1.00s its angular momentum is 5.00 kg m² /swhat its angular momentum is at t =3.00s

Short answer

Answer

The angular momentum of the disk at t = 3.00 is$\text{L}=23{\text{kgm}}^2/\text{s}$

Step-by-step solution

Given

1. The moment of inertia of disk,$\text{I}=7.00{\text{kgm}}^2.$
2. The torque of the rotating disk,$\text{T}=\left(5.00+2.00\text{t}\right)\text{Nm}.$

To understand the concept

Using the relation between torque and angular momentum, we can find the angular momentum from torque.

Formula:

$\mathit{\tau }\mathbf{=}\frac{\mathbf{d}\mathbf{L}}{\mathbf{d}\mathbf{t}}$

Calculate the angular momentum as a function of time

We know that,

$T=\frac{dL}{dt}$

Therefore, the angular momentum as a function of time is

$$\begin{gathered} \text{L}\left(\text{t}\right)=\int Tdt \\ \Rightarrow \text{L}\left(\text{t}\right)=\int (5+2\text{t})dt \\ \Rightarrow \text{L}\left(\text{t}\right)=5\text{t}+\frac{2{\text{t}}^2}{2}+\text{c} \\ \Rightarrow \text{L}\left(\text{t}\right)=5\text{t}+{\text{t}}^2+\text{c} \end{gathered}$$

(Where c is constant of integration)

We are given that$\text{L}=5\text{att}=1\text{s}.$

$$\begin{gathered} \Rightarrow L\left(\text{t}\right)=5\text{t}+{\text{t}}^2+\text{c} \\ \Rightarrow 5=5\left(1\right)+{\left(1\right)}^2+\text{c} \\ \Rightarrow 5=6+\text{c} \\ \Rightarrow \text{c}=-1 \end{gathered}$$

Therefore, angular momentum is

$\text{L}\left(\text{t}\right)=5\text{t}+{\text{t}}^2-1$

Calculate the angular momentum at t =3.00 s

Therefore, the angular momentum of the disk at t =3.0 s is

$$\begin{gathered} \text{L}\left(\text{t}=3\right)=5\left(3\right)+{\left(3\right)}^2-1 \\ \Rightarrow \text{L}=15+9-1 \\ \Rightarrow \text{L}=23{\text{kgm}}^2/\text{s} \end{gathered}$$

Angular momentum of the disk at t =3.00s is$23{\text{kgm}}^2/\text{s}$