Question

Darcel and Chandra are sitting in the park after physics class, and they notice some children rolling various objects down a slight incline in the sidewalk. Darcel is curious how the masses and shapes of the objects affect their motion, and Chandra says that it might be fun to think about the energy of motion associated with rolling objects. The friends decide to use the simulation to explore the motion and energies of various objects rolling down an incline.

They can click on an object (solid sphere, spherical shell, hoop, or cylinder) to position it at the top of the incline and click "roll" to release it. The translational and rotational speeds of the object and the time interval the object is on the incline are shown, together with a graph of the percentages of the different energy types as the object rolls down the incline. The "roll" button turns into a "pause" button while the object is in motion to allow them to examine different parameters at various points. For the object–Earth system in the simulation, the zero configuration of gravitational potential energy is defined to occur when the object is at the bottom of the incline. For the simulation, air resistance and any rolling friction effects are neglected, and the masses of all the objects are equal.

translational speed 0.00 % 0 0 0 0 5 100 10 m/s rotational speed 0.00 rad/s time on ramp 0.00 translational potential & rotational energy kinetic energy total energy roll h-14 m hoops cylinders spheres (solid) spheres (hollow)

Now Chandra and Darcel decide to try a problem.

Suppose that the height of the incline is h = 14.9 m. Find the speed at the bottom for each of the following objects.

solid sphere	_____m/s
spherical shell	_____m/s
hoop	_____m/s
cylinder	_____m/s

In a race, which object would win?

-- solid sphere

-- spherical shell    

-- hoop

-- cylinder

-- tie

Answer 1

solid sphere would win.

v_solidsphere = 14.4 m/s
v_sphericalshell = 13.2 m/s
v_hoop = 12.1 m/s
v_cyllinder = 13.9 m/s

for solid sphere,

Apply conservation of energy

final 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

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

v_solidsphere = sqrt(10*g*h/7)

= sqrt(10*9.8*14.9/7)

= 14.4 m/s

for sphericalshell,

Apply conservation of energy

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

v_sphericalshell = sqrt(6*g*h/5)

= sqrt(6*9.8*14.9/5)

= 13.2 m/s

for hoop,

Apply conservation of energy

final 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)*m*r^2*w^2 = m*g*h

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

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

m*v^2 = m*g*h

v_hoop = sqrt(g*h)

= sqrt(9.8*14.9)

= 12.1 m/s

for cyllinder,

Apply conservation of energy

final 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)*m*r^2*w^2 = m*g*h

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

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

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

v_cyllinder = sqrt(4*g*h/3)

= sqrt(4*9.8*14.9/3)

= 13.9 m/s

clearly,

v_hoop < v_spericalshell < v_cyllinder < v_soildsphere