Question

You are at the park with your little brother, when you notice a small merry-go-round with a radius that looks to be about 1.5 m. Your brother climbs on, and you give him a spin. From rest, you speed up smoothly, completing one full lap in 3.0 seconds. You wonder what sort of acceleration (magnitude and direction) your brother experiences at the very start of the motion, and at the very end of the first lap. He is quite young and hasn't studied any physics, so he does not know how to answer your question. You decide to give him your smartphone (which has a built-in accelerometer), and repeat the experiment. What will the smartphone record

Answer 1

Answer:

Explanation:

angle covered in one rotation = 2π radian

θ = ωt + 1/2 αt²

θ is angle rotated in time t with initial angular velocity of ω and angular acceleration α .

Putting the values

2π = 0 + 1/2 x α x 3²

α = 1. 4 radian / s²

linear acceleration =  α x r = 1.4 x 1.5 = 2.1 m / s².

Initial acceleration = 2.1 m /s²

final angular velocity = α t = 1.4 x 3 = 4.2 radian / s

linear velocity = 4.2 x 1.5 = 6.3 m /s

centripetal acceleration = v² / R = 6.3² / 1.5 = 26.46 m /s²

radian acceleration = 26.46 m /s

tangential acceleration = 2.1 m /s²

Total final acceleration = √ ( 26.46² + 2.1² )

= √ ( 700.13 + 4.41)

Final acceleration = 26.53 m / s²

Answer 2

The kid will undergo a circular motion and the smartphone which has a built-in accelerometer will record the acceleration $26.54 \,m/s^2$, which is the magnitude of the vector sum of both radial acceleration and tangential acceleration.

Circular Motion

The angle covered in one full rotation, i.e.; the angular displacement is,

$\theta = 2\pi\,\, radian$

The kinematics equation in the case of rotational motion is given by;

$\theta = \omega_i \,t + \frac{1}{2} \alpha t^2$

But the initial angular velocity, $\omega_i = 0$

Time taken to complete one lap is, $t = 3\,s$

Therefore, substituting the given values in the kinematics equation;

$2\pi \;rad=0 + \frac{1}{2} \alpha\times (3)^2$

$\implies \alpha = \frac{4 \pi\,\,rad}{9\,s^2} =1.395\,rad/s^2 \approx 1.4\,rad/s^2$

We know that linear acceleration (here tangential direction) is given by,

$a_T=a = r \alpha = 1.5\,m \times 1.4\; rad/s^2=2.1\,m/s^2$

Also, the final angular velocity after 3 s is given by;

$\omega_f = \alpha t=1.4\,rad/s^2 \times 3\,s = 4.2\,rad/s$

Therefore, the final linear velocity is;

$v_f = r \omega = 1.5\,m \times 4.2\,rad/s = 6.3\,m/s$

We know that the centripetal or the radial acceleration is given by;

$a_r = \frac{v^2}{r} = \frac{(6.3\,m/s)^2}{1.5m}=26.46\,m/s^2$

Therefore, the total acceleration is given by;

$a_{total} = \sqrt{(a_T)^2 + (a_r)^2\,} =\sqrt{(2.1)^2 + (26.46)^2} =\sqrt{704.54} =26.54 \,m/s^2$

Learn more about circular motion here:

https://brainly.com/question/2562955