Question

a uniform sphere of mass 2.5 kg and radius 0.4 m rolls down a ramp inclined at an angle 0.13 radians to the horizontal. what is the acceleration of the rolling sphere in m/s2 ?

Answer 1

The acceleration of the rolling sphere is 1.96 m/s^2.

Step-by-step explanation:

To find the acceleration of the rolling sphere, you can use the equation:
acceleration = (sin(theta) * g) / (1 + (k^2 * R^2) / (2/5 * m * R^2)),
( where theta is the angle of the ramp, g is the acceleration due to gravity (9.8 m/s^2), R is the radius of the sphere, k is the coefficient of rolling resistance, and m is the mass of the sphere).

Plugging in the values given:
theta = 0.13 radians
R = 0.4 m
m = 2.5 kg

Assuming that the ramp is rough and there is no slipping, the coefficient of rolling resistance (k) is 0.3.

Calculating the acceleration:
acceleration = (sin(0.13) * 9.8) / (1 + (0.3^2 * 0.4^2) / (2/5 * 2.5 * 0.4^2))

= 1.96 m/s^2