What is the magnitude of torque acting on a particle moving in the xy - plane about the origin if its angular momentum is `4.0sqrt(t)kgm^2s^-1` ?

Options

A. `8t^(3//2)`
B. `4/(sqrt(t))`
C. `2/(sqrt(t))`
D. `3/(2sqrt(t))`

Answer

To find the magnitude of the torque acting on a particle moving in the xy-plane about the origin, given its angular momentum \( L = 4.0 \sqrt{t} \, \text{kg m}^2/\text{s} \), we can follow these steps:

### Step 1: Understand the relationship between torque and angular momentum
Torque (\( \tau \)) is defined as the rate of change of angular momentum (\( L \)):
\[ \tau = \frac{dL}{dt} \]

### Step 2: Differentiate the angular momentum with respect to time
Given \( L = 4.0 \sqrt{t} \), we need to differentiate this with respect to \( t \):
\[ \frac{dL}{dt} = \frac{d}{dt}(4.0 \sqrt{t}) \]

### Step 3: Apply the differentiation rule
Using the power rule for differentiation, where \( \sqrt{t} = t^{1/2} \):
\[ \frac{dL}{dt} = 4.0 \cdot \frac{1}{2} t^{-1/2} \cdot \frac{dt}{dt} = 4.0 \cdot \frac{1}{2} t^{-1/2} = 2.0 t^{-1/2} \]

### Step 4: Simplify the expression for torque
Now we can express the torque:
\[ \tau = \frac{dL}{dt} = 2.0 t^{-1/2} \]
This can also be written as:
\[ \tau = \frac{2.0}{\sqrt{t}} \]

### Step 5: Finalize the expression for torque
Thus, the magnitude of the torque acting on the particle is:
\[ \tau = 2 \sqrt{t} \, \text{N m} \]