Question

a wheel of radius 1m is spinning with angular velocity which varies with time as 5t^2 rad/s. The acceleration of the point on the wheel's rim at t = 1 s, has the magnitude equal to (i)5 root 29 m/s^2 (ii) 25 m/s^2 (iii) 10m/s^2 (iv) 29 root 5 m/s^2
PLEASE EXPLAIN COMPLETELY

Answer 1

Answer is given in the pic.

And don't forget to solve my question.

Answer 2

Answer: Acceleration of the point is $\bold{5\sqrt{29} \ m/s^2}$.

Given: Radius 1 m and angular velocity varies with time is 5t^2 rad/s.

To Find: The acceleration of the point on the wheel's rim.

Step-by-step explanation:

Step 1: A pseudovector used in physics to express how quickly the angular location or orientation of an item changes over time is called an angular velocity or rotational velocity (ω or Ω). (i.e. how quickly an object rotates or revolves relative to a point or axis). The pseudovector's direction is normal to the instantaneous plane of rotation or angular displacement, and its magnitude corresponds to the object's angular speed, or the rate at which it rotates or revolves. The right-hand rule is typically used to specify the direction of angular motion.

Step 2: Angular velocity ($\omega$) = $5t^2$

Velocity at rim

$v=r\omega\\\\v=1\times 5t^2\\\\v=5t^2$

Tangential acceleration = $a_t=\frac{dv}{dt} =10t$

At $t= 1 sec$, $a_t=10$

Centripetal acceleration = $a_c=\frac{v^2}{r} =25t^4$

At $t= 1 sec$,  $a_c=25$

Step 3: Now,

$a_{net}=\sqrt{{a_c}^2+{a_t}^2} \\\\a_{net}=\sqrt{{25}^2+{10}^2} \\\\a_{net}=\sqrt{625+100} \\\\a_{net}=\sqrt{725} \\\\a_{net}=5\sqrt{29}$

Hence correct answer is $5\sqrt{29} \ m/s^2$.

For more questions please follow the link given below.

https://brainly.in/question/50268315

https://brainly.in/question/49928587

#SPJ2