Question

An object moves with a constant speed in a horizontal circular motion. If the radius of the circle is 5 m and the period of revolution is 2.5 s, find the acceleration, in m/s2, towards the centre of the circle.

Answer 1

Answer: The acceleration towards the center of the circle is approximately 31.583 m/s^2.

Step-by-step explanation:

In circular motion, the centripetal acceleration is the acceleration towards the center of the circle that keeps the object moving in a circular path. The centripetal acceleration (a_c) can be calculated using the following formula:

a_c = (4 * π^2 * r) / T^2

where:

a_c is the centripetal acceleration (in m/s^2).

π (pi) is a mathematical constant approximately equal to 3.14159.

r is the radius of the circle (in meters).

T is the period of revolution (in seconds).

Given that the radius of the circle (r) is 5 m and the period of revolution (T) is 2.5 s, we can calculate the centripetal acceleration as follows:

a_c = (4 * π^2 * 5) / (2.5^2)

a_c = (4 * 3.14159^2 * 5) / 6.25

a_c = (4 * 9.8696 * 5) / 6.25

a_c = (39.4784 * 5) / 6.25

a_c = 197.392 / 6.25

a_c ≈ 31.583 m/s^2

So, the acceleration towards the center of the circle is approximately 31.583 m/s^2.