A uniform hollow spherical ball of mass 1.75 kg and radius 40.0 cm rolls without

Question

A uniform hollow spherical ball of mass 1.75 kg and radius 40.0 cm rolls without slipping up a ramp that rises at 30.0 degrees above the horizontal. The speed of the ball at the base of the ramp is 2.63 m/s. (a) While the ball is moving up the ramp, find the magnitude of the acceleration of its center of mass. (b) While the ball is moving up the ramp, find the magnitude of the friction force acting on it due to the surface of the ramp. Please give exact answer, don't copy and paste generic formula already online. Tried the equations floating online and they didn't work.

Explanation / Answer

a)
let h is the vertical height raised by the hollow sphere.

Apply conservation of energy

initial kinetic energy = final potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(2/3)*m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/3)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/3)*m*v^2 = m*g*h

(5/6)*v^2 = g*h

==> h = 5*v^2/(6*g)

= 5*2.63^2/(6*9.8)

= 0.588 m

distance travelled along the ramp, d = h/sin(30)

= 0.588/sin(30)

= 1.176 m

let a is the acceleation of the sphere,

now use, (vf^2 - vi^2) = 2*a*d

==> a = (vf^2 - vi^2)/(2*d)

= (0^2 - 2.63^2)/(2*1.176)

= -2.94 m/s^2 <<<<<<<<<<<<----------------------Answer

b) net force acting along the ramp,

Fnet = -m*g*sin(30) + F_friction

m*a = -m*g*sin(30) + F_friction

F_friction = m*a + m*g*sin(30)

= 1.75*(-2.94) + 1.75*9.8*sin(30)

= 3.43 N <<<<<<<<<--------------------------Answer

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