Question

Rolling Without Slipping & Conservation of Mechanical Energy 1. A caveman applies a horizontal force of 800 N at height of 0.1 m above the center of a large spherical boulder of mass 400 kg and radius of 0.5 m (treat as a sphere: lon-2/5 mr2). Assume the sphere starts from rest and rolls horizontally without slipping. a) Draw a free-body diagram and label all the forces at their point of contact. b) Write equations applying Newton's 2 nd law for rotation and translation for the boulder. c) Solve for the translational and angular acceleration of the boulder. d) What is the minimum coefficient of static friction to prevent slipping? e) If the boulder were replaced with a hollow hoop of the same mass and radius (assume no slipping), how would that have affected i) the angular acceleration ii) the linear acceleration ili) the static friction needed not to slip

Answer 1

a] For the free body diagram, the force by the caveman will be horizontal and it will be at a distance of 0.6 m from the ground.

the normal force R will be vertically upwards with magnitude 400x9.8 =3920 N

weight mg will be acting vertically downwards with magnitude 3920 N.

b]

for translational acceleration,

a = F/m = 800/400 = 2 m/s²

for rotational acceleration,

$\alpha = \frac{\tau}{I}$

where the numerator is the applied torque and the denominator is the moment of inertia.

=> $\alpha = \frac{(0.1)800}{\frac{2}{5}400(0.5)^2} = 2 rad/s^2$

d]

To prevent slipping, the net force on the system should be zero

so,

F - u_sR = 0

800 - u_s(3920) = 0

=> u_s = 800/3920 = 0.2041

this is the minimum coefficient of static friction required to prevent slipping.

e] If the boulder is replaced with hollow hoop of same mass and radius, the angular acceleration will change since the new moment of inertia will be: I = mr²

so, the angular acceleration will decrease [since moment of inertia was in the denominator]

the linear acceleration will be unaffected since it is independent of the moment of inertia.

the static friction will also remain the same since it was dependent on the applied force and the normal reaction.