A body moves along a circular path with uniform angular speed of 0.6 rad s-1 and at a constant speed of 3.0ms-1. Calculate the acceleration of the body towa...

Assessment: WAEC SSCE - Physics - 1993 (Objective) Subject: Physics

Question 1 Report

A body moves along a circular path with uniform angular speed of 0.6 rad s-1 and at a constant speed of 3.0ms-1. Calculate the acceleration of the body towards the centre of the circle.
Answer Details
When an object moves in a circle, it is constantly changing its direction, which means that it is accelerating. This is because acceleration is the rate of change of velocity, and velocity is a vector that includes both speed and direction. In this case, the object is moving with a uniform angular speed, which means that it is moving at a constant speed in a circle. The formula for acceleration towards the center of a circle is given by a = v^2 / r, where v is the speed of the object and r is the radius of the circle. We know that the speed of the object is 3.0 m/s, and we are given the angular speed of the object as 0.6 rad/s. Angular speed is the rate at which the object moves around the circle in radians per second. To find the radius of the circle, we need to use the relationship between angular speed, linear speed, and radius, which is given by v = ωr, where ω is the angular speed. Rearranging this formula gives us r = v/ω. Substituting the values we have, we get r = 3.0 m/s / 0.6 rad/s = 5.0 m. Now we can calculate the acceleration towards the center of the circle using the formula a = v^2 / r. Substituting the values we have, we get a = (3.0 m/s)^2 / 5.0 m = 1.8 m/s^2. Therefore, the answer is option D: 1.8 ms^-2.

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