How much force will it exert on the steel plate?

Question

Assume a steel cylinder is placed at the top of the ramp. At the end of the ramp a steel plate is placed to stop the cylinder. If it begins to roll on the ramp then How much force will it exert on the steel plate? How much speed will it get at the end of the Ramp? Please provide the formula to calculate and explain me in great detail.

Cylinder Dia – 15 cm

Cylinder Weight – 10 Kg

Cylinder Length – 10 cm ( Contact Surface)

Ramp length – 60 cm

Ramp Angle – 15 degree.

Answer 1

if there is no friction there will be no torque to start the cylinder rolling. Regardless of the angle, the cylinder will slide down the incline in unconstrained motion.

I would start with the potential energy of the system

E=mgh=potential energy=kinetic energy at the bottom

so h=sin(15)*.6=.155291427062 meters

m is 10 kg

and g is 9.81 m/s^2

total energy is 15.2340889947 Joules

m*Ao=m*g*sin(angle)

m's cancel

x is parallel to the inclined plane surface

Ao,x=9.81*sin(15)= 2.53901483246 m/s^2

Force=m*Ao,x=25.3901483 Newtons

the speed would be v=sqrt(2*g*h)

v=sqrt(2*9.81*.155291427062 )=1.7455136 m/s

if the cylinder is rolling then there must be friction

the equations becomes more difficult

mass*gravity*sin(angle)-Force of friction=m*Ao,x

and the second equation is Force of friction*radius=Moment of inertia*rotational acceleration

the equation would be

10*9.81*sin(15)- Ffriction=10*Ao,x

Ffriction*.075=.5*10*.075^2*angular acceleration

if you know the material of the ramp then you could solve this equation. Ffriction=fmaterial*Normal force