Answer to Question #180589 in Optics for ankit

Question

A merry-go-round is initially at rest. On being given a constant angular acceleration it reaches an angular speed of 1srad50.0in 10.0 s. At t=10.0 s, determine the magnitude of: (i) the angular acceleration of the merry-go-round ; (ii) the linear velocity of a child sitting on the merry-go-round at a distance of3.0 m from its centre; (iii) the tangential acceleration of the child; (iv) the centripetal acceleration of the child; and (v) the netacceleration of the child.

Answer 1

i) By the definition of the angular acceleration, we have:

$$\alpha=\dfrac{\Delta \omega}{\Delta t}=\dfrac{10.5\ \dfrac{rad}{s}}{10.0\ s}=1.05\ \dfrac{rad}{s^2}.$$

ii) We can find the linear velocity of the child sitting on the merry-go-round at a distance of 3.0 m from its center from the formula:

$$v=\omega r=10.5\ \dfrac{rad}{s}\cdot 3.0\ m=31.5\ \dfrac{m}{s}.$$

iii) We can find the tangential acceleration of the child from the formula:

$$a_t=r\alpha=3.0\ m\cdot1.05\ \dfrac{rad}{s^2}=3.15\ \dfrac{m}{s^2}.$$

iv) We can find the centripetal acceleration of the child from the formula:

$$a_c=\dfrac{v^2}{r}=\dfrac{(31.5\ \dfrac{m}{s})^2}{3.0\ m}=330.75\ \dfrac{m}{s^2}.$$

v) We can find the net acceleration of the child from the Pythagorean theorem:

$$a_{net}=\sqrt{a_t^2+a_c^2},$$ $$a_{net}=\sqrt{(3.15\ \dfrac{m}{s^2})^2+(330.75\ \dfrac{m}{s^2})^2}=330.76\ \dfrac{m}{s^2}.$$