A solid cylinder and another solid cylinder with the same mass but double the radius start at the same height on an incline plane with height h and roll without slipping. Consider the cylinders as disks with moment of inertias I=(1/2)mr^2. Which one reaches the bottom of the incline plane first?
According to this, the velocity of any body rolling down the plane is
v=(2 g h/1 + c) ^½
where c is the constant in moment of inertia (for example, c=2/5 for a solid sphere).
My thought process was that since the radius doubled, c=2. So, the velocity of the doubled cylinder would be less, therefore finishing later. Similarly, if it’s moment of inertia increases, it’s angular and linear acceleration decreases. However, my other peers and even my professor disagree, saying that radius and mass do not play a role in the velocity of the body, since both m and r will cancel in an actual calculation of the velocity.
Could anyone elaborate on whether I am right or wrong?
Best Answer
The following equation from @R. Romero's analysis is correct: $$Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I\left(\frac{v^2}{R^2}\right)\tag{1}$$But, the moment of inertia of a cylinder is given by: $$I=M\frac{R^2}{2}\tag{2}$$ So, combining Eqns. 1 and 2 gives:$$Mgh=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2\tag{3}$$Calcelling M from both sides of the equation yields:$$gh=\frac{1}{2}v^2+\frac{1}{4}v^2\tag{4}$$So, solving for v, we have:$$v=\sqrt{\frac{4gh}{3}}$$Note that this is independent of the radius of the cylinder. So, both cylinders roll down the ramp in the same amount of time.