The angular momentum of a body is given by `L=5t^2+2t+1kg m^2/s`.The torque acting on the body at `t=1`s is

Options

A. `12N-m`
B. 10N-m
C. 4N-m
D. Zero

Answer

To find the torque acting on the body at \( t = 1 \) second, we start with the given expression for angular momentum \( L \):

\[ L = 5t^2 + 2t + 1 \, \text{kg m}^2/\text{s} \]

### Step 1: Differentiate the angular momentum with respect to time
The torque \( \tau \) is defined as the rate of change of angular momentum with respect to time:

\[ \tau = \frac{dL}{dt} \]

Now we differentiate \( L \):

\[ \frac{dL}{dt} = \frac{d}{dt}(5t^2 + 2t + 1) \]

### Step 2: Apply the differentiation rules
Using the power rule for differentiation, we differentiate each term:

- The derivative of \( 5t^2 \) is \( 10t \).
- The derivative of \( 2t \) is \( 2 \).
- The derivative of the constant \( 1 \) is \( 0 \).

Thus, we have:

\[ \frac{dL}{dt} = 10t + 2 \]

### Step 3: Substitute \( t = 1 \) second into the derivative
Now we need to find the torque at \( t = 1 \) second:

\[ \tau = 10(1) + 2 \]

Calculating this gives:

\[ \tau = 10 + 2 = 12 \, \text{N m} \]

### Final Answer
The torque acting on the body at \( t = 1 \) second is:

\[ \tau = 12 \, \text{N m} \]

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