A fan of moment of inertia 0.6 , kg cdot m^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The angular momentum $L$ of a rotating body is given by the formula $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular v

Vedclass · Accepted Answer

$0.6\pi \, kg \cdot m^2/s$

Vedclass · Answer

(A) The angular momentum $L$ of a rotating body is given by the formula $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.Given: Moment of inertia $I = 0.6 \, kg \cdot m^2$.Frequency of rotation $f = 0.5 \, rev/s$.The angular velocity $\omega$ is related to the frequency by $\omega = 2\pi f$.Substituting the values: $\omega = 2 	imes \pi 	imes 0.5 = \pi \, rad/s$.Now,calculate the angular momentum: $L = I\omega = 0.6 	imes \pi = 0.6\pi \, kg \cdot m^2/s$.Therefore,the correct option is $A$.

$A$ fan of moment of inertia $0.6 \, kg \cdot m^2$ is turned up to a working speed of $0.5$ revolutions per second. The angular momentum of the fan is:

Similar Questions

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $X$-axis. The locus of the other end of the diameter through $A$ is

If ${x^y} = {y^x},$ then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are two functions such that $g \circ f: A \rightarrow C$ is a bijection,then which one of the following is always true?

Observe the statements given below :
Assertion $(A)$ : $f(x)=x e^{-x}$ has the maximum at $x=1$
Reason $(R)$ : $f^{\prime}(1)=0$ and $f^{\prime \prime}(1) < 0$
Which of the following is correct?

For all real $x$,the minimum value of the function $f(x)=\frac{1-x+x^2}{1+x+x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module