Question

A child is riding a merry-go-round, which has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2. the child is standing 4.65 m from the center of the merry-go-round. what angle does the acceleration of the child make with the tangential direction

Answer 1

To find the angle that the acceleration of the child makes with the tangential direction, we can use trigonometry. Using the formulas for tangential acceleration and radial acceleration, we can calculate the respective values. Finally, we can use the formula for the angle to find the answer.

Step-by-step explanation:

To find the angle that the acceleration of the child makes with the tangential direction, we need to use trigonometry. We can use the formula:

tan(theta) = at/ar

where theta is the angle, at is the tangential acceleration, and ar is the radial acceleration. The tangential acceleration can be found using the formula:

at = r * alpha

where r is the distance from the center of the merry-go-round and alpha is the angular acceleration. Plugging in the values, we have:

at = 4.65 m * 0.745 rad/s2 = 3.45 m/s2

The radial acceleration is given by:

ar = r * omega2

where omega is the angular velocity. Plugging in the values, we have:

ar = 4.65 m * (1.25 rad/s)2 = 7.27 m/s2

Now we can find the angle using the formula:

theta = atan(at/ar)

Plugging in the values, we have:

theta = atan(3.45/7.27) = 0.45 radians

Therefore, the angle that the acceleration of the child makes with the tangential direction is 0.45 radians.

Answer 2

w = instantaneous angular speed = 1.25 rad/s

r = radius = 4.65 m

α = angular acceleration = 0.745 rad/s²

centripetal acceleration is given as

`a_(c)` = rw²

`a_(c)` = (4.65 ) (1.25)² = 7.27 m/s²

tangential acceleration is given as

`a_(t)` = rα

`a_(t)` =4.65 x 0.745 = 3.46 m/s²

angle is given as

`\theta = tan^(-1)((a_(c))/(a_(t)))`

`\theta = tan^(-1)((7.27)/(3.46))`

`\theta` = 64.5 deg