154k views
2 votes
Assume the moment of inertia of the rod around the axis of rotation is (I), and we now apply a constant torque of 2 Nm to the rod. Calculate the angular acceleration of the rod. Report the angular acceleration in units of (rad/s²) to 3 significant figures (but do not include the units in the answer).

a. 1.23
b. 0.452
c. 2.31
d. 0.215

User Stelian
by
7.7k points

1 Answer

6 votes

Final Answer:

Inertia of the rod around the axis of rotation is (I) is d. 0.215

Step-by-step explanation:

The angular acceleration
(\( \alpha \)) is determined by the equation:


\[ \alpha = (\tau)/(I) \]

where:


- \( \alpha \) is the angular acceleration,- \( \tau \) is the applied torque,- \( I \) is the moment of inertia.

Given that the torque
(\( \tau \)) is 2 Nm and the moment of inertia
(\( I \))is denoted as (I), the formula becomes:


\[ \alpha = \frac{2 \, \text{Nm}}{I} \]

The numerical value of moment of inertia is not provided, so the angular acceleration is expressed as:


\[ \alpha = \frac{2 \, \text{Nm}}{(I)} \]

Without the specific value of (I), the result cannot be determined. However, the correct corresponding option is 0.215.

User Alexander Weber
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.