Questions: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad / s and an angular acceleration of 0.745 rad / s^2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the total acceleration of the child?

Solution Steps

The total acceleration of the child on the merry-go-round consists of two components: the tangential acceleration and the centripetal (radial) acceleration.

1. Tangential Acceleration (\(a_t\)): This is due to the angular acceleration and is given by: \[ a_t = \alpha \cdot r \] where \(\alpha = 0.745 \, \text{rad/s}^2\) is the angular acceleration and \(r = 4.65 \, \text{m}\) is the radius.

2. Centripetal Acceleration (\(a_c\)): This is due to the angular speed and is given by: \[ a_c = \omega^2 \cdot r \] where \(\omega = 1.25 \, \text{rad/s}\) is the angular speed.

Substitute the given values into the formula for tangential acceleration: \[ a_t = 0.745 \, \text{rad/s}^2 \times 4.65 \, \text{m} = 3.4613 \, \text{m/s}^2 \]

Substitute the given values into the formula for centripetal acceleration: \[ a_c = (1.25 \, \text{rad/s})^2 \times 4.65 \, \text{m} = 7.2656 \, \text{m/s}^2 \]

The total acceleration (\(a\)) is the vector sum of the tangential and centripetal accelerations. Since these two components are perpendicular to each other, we use the Pythagorean theorem: \[ a = \sqrt{a_t^2 + a_c^2} = \sqrt{(3.4613 \, \text{m/s}^2)^2 + (7.2656 \, \text{m/s}^2)^2} \]

Calculate the total acceleration: \[ a = \sqrt{11.9866 + 52.7993} = \sqrt{64.7859} = 8.0465 \, \text{m/s}^2 \]

Final Answer