Question

if a sphere rolls in 5.3 sec along a plane 1m in length of which the upper end is raised 0.01m above the lower,find acceleration due to gravity

Answer 1

    We have an inclined plane on which a sphere of mass M and radius R rolls down in 5.3 sec. Since the sphere rolls down without slipping, there is friction force f on the inclined plane. Friction force f acts upwards to prevent slipping.

Moment of Inertia I = 2/5 M R²
Angular acceleration = α
Linear acceleration down the incline = a = R α

Torque acting on the sphere about the center
    T = f R = I α = 2/5 * M R² α = 2/5 M R a
     f = 2/5 M a

Given the angle of incline = Ф.    Sin Ф = 0.01m/1m = 0.01

The component of gravity along the plane = g SinФ = 0.01 g

Equation of motion for the sphere:
          m g sinФ - f = m a
          m g SinФ = f + m a = 7/5 m a
          a = 5/7 * g SinФ = 0.05 g / 7 

Using equations of kinematics:   (t = 5.3 sec given)
          s = u t + 1/2 a t²
          1 m = 0 + 1/2 * 0.05 g / 7 * 5.3²
           g = 14/1.4045 = 9.968 m/sec²

 g = 9.968 m/sec² or nearly 10 m/sec²

Answer 2

your answer is 10 m/ s2 mate