An athlete throws a discus from rest to a fin | vedclass

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

Question

(A) Given: Initial angular velocity $\omega_{0} = 0 \, rad \, s^{-1}$,final angular velocity $\omega = 15 \, rad \, s^{-1}$,time $t = 0.270 \, s$,and

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) Given: Initial angular velocity $\omega_{0} = 0 \, rad \, s^{-1}$,final angular velocity $\omega = 15 \, rad \, s^{-1}$,time $t = 0.270 \, s$,and radius $r = 0.810 \, m$.First,we calculate the angular acceleration $\alpha$ using the formula $\omega = \omega_{0} + \alpha t$.Since $\omega_{0} = 0$,we have $\alpha = \frac{\omega}{t} = \frac{15}{0.270} \, rad \, s^{-2}$.The tangential acceleration $a_{t}$ is given by $a_{t} = r \alpha$.Substituting the values: $a_{t} = 0.810 	imes \frac{15}{0.270} = 0.810 	imes 55.55 = 45 \, m \, s^{-2}$.Thus,the acceleration of the discus is $45 \, m \, s^{-2}$.

An athlete throws a discus from rest to a final angular velocity of $15 \, rad \, s^{-1}$ in $0.270 \, s$ before releasing it. During acceleration,the discus moves along a circular arc of radius $0.810 \, m$. The acceleration of the discus before it is released is ....... $m \, s^{-2}$.

Similar Questions

$A$ flywheel rotates with a uniform angular acceleration. Its angular velocity increases from $20\pi \, rad/s$ to $40\pi \, rad/s$ in $10 \, s$. How many rotations did it make in this period?

$A$ particle starts rotating from rest. Its angular displacement is expressed by the following equation $\theta = 0.025t^2 - 0.1t$,where $\theta$ is in radians and $t$ is in seconds. The angular acceleration of the particle is:

If the position vector of a particle is $\vec{r} = (3\hat{i} + 4\hat{j}) \text{ m}$ and its angular velocity is $\vec{\omega} = (\hat{j} + 2\hat{k}) \text{ rad/s}$, then its linear velocity $\vec{v}$ is (in $\text{m/s}$):

$A$ wheel has a diameter of $1 \ m$. If it makes $30$ revolutions per second,then the linear speed of a point on its circumference will be:

$A$ wheel undergoes a constant angular acceleration from time $t=0$ to $t=20 \ s$ and thereafter angular acceleration is zero. If angular velocity at $t=2 \ s$ is found to be $5 \ rad/s$, then the number of revolutions made by the wheel in the time interval $t=0 \ s$ to $t=50 \ s$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module