Question

Any help with detail explanation is appreciated. Thank you!

13) (SLO 4) (20 points) A uniform marble of mass M rolls down a symmetric bowl, starting from rest at the top of the left side. The top each side is a distance h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping. The right half has no friction as it is coated with oil and smooth. The bowl is rough enough to cause the marble to roll such that Vcm-Rw, where R is the radius of the marble

a. What is the speed of the marble of the bottom of the bowl. Know that I=2/5mr^2

b. Similiar to a. What is the speed of the disk of the bottom of the bowl. Know that I=1/2mr^2

Answer 1

a) let v is the linear speed and w is the angular speed of the marble at the bottom.

Apply conservation of mecahnical energy

final translational kinetic energy + rotational kinetic energy = initial potential energy

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

(1/2)*m*v^2 + (1/2)*(2/5)*m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*v^2 = m*g*h (since v = r*w)

(7/10)*m*v^2 = m*g*h

(7/10)*v^2 = g*h

v^2 = 10*g*h/7

v = sqrt(10*g*h/7) <<<<<<<------------------Answer

b)

let v is the linear speed and w is the angular speed of the marble at the bottom.

Apply conservation of mecahnical energy

final translational kinetic energy + rotational kinetic energy = initial potential energy

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

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

(1/2)*m*v^2 + (1/4)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/4)*m*v^2 = m*g*h (since v = r*w)

(3/4)*m*v^2 = m*g*h

(3/4)*v^2 = g*h

v^2 = 4*g*h/3

v = sqrt(4*g*h/3) <<<<<<<------------------Answer