Double A
Nov11-04, 04:55 PM
I have been working on this problem for the past 48 hours and out off all the ways I keep doing it I still get the wrong answer. Maybe someone can show me what I did wrong in the following calculations.
The Problem:
Many machines employ cams fro various purposes, such as opening and closing valves. In Figure P10.29 (Not shown here), the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed \omega about the axis of the shaft?
My closest answer:
To find the moment of inertia of the whole cylinder I have to find the total mass of the cylinder because M is the mass of the cam with the hole in it. So,
M_d = M - \frac{1}{4}M=\frac{3}{4}M
M_d=\frac{3}{4}M
M=\frac{4}{3}M_d
Now, the moment of inertia about the center of mass of the disk is
I_{CM_D} = \frac{1}{2}MR^2+MD^2
The parallel axis theorem is needed because the disk is rotating about an axis not at its center of mass.
I_{CM_D} = \frac{1}{2}(\frac{4}{3}M_d)R^2+(\frac{1}{4}M)(\fra c{R}{2})^2 = \frac{35}{48}MR^2
The moment of inertia of the hole drilled into the cam is
I_{hole} = I_{CM} + MD^2 = \frac{1}{2}(\frac{M}{4})R^2+(\frac{M}{4})R^2= \frac{3}{16}MR^2
The total moment of inertia of the cam is
I_{total} = I_{disk} - I_{hole} = \frac{35}{48}MR^2 - \frac{3}{16}MR^2 = \frac{13}{24}MR^2
The equation for rotational kinetic energy is
K_R = \frac{1}{2}I\omega^2
So then the kinetic energy of the cam is
K_R = \frac{1}{2}(\frac{13}{24}MR^2)\omega^2 = \frac{13}{48}MR^2\omega^2
However, when I reference my answer with the true answer I am wrong. The true answer is
K_R = \frac{23}{48}MR^2\omega^2
I'm off by \frac{5}{24}. Some one please tell me where I went wrong.
The Problem:
Many machines employ cams fro various purposes, such as opening and closing valves. In Figure P10.29 (Not shown here), the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed \omega about the axis of the shaft?
My closest answer:
To find the moment of inertia of the whole cylinder I have to find the total mass of the cylinder because M is the mass of the cam with the hole in it. So,
M_d = M - \frac{1}{4}M=\frac{3}{4}M
M_d=\frac{3}{4}M
M=\frac{4}{3}M_d
Now, the moment of inertia about the center of mass of the disk is
I_{CM_D} = \frac{1}{2}MR^2+MD^2
The parallel axis theorem is needed because the disk is rotating about an axis not at its center of mass.
I_{CM_D} = \frac{1}{2}(\frac{4}{3}M_d)R^2+(\frac{1}{4}M)(\fra c{R}{2})^2 = \frac{35}{48}MR^2
The moment of inertia of the hole drilled into the cam is
I_{hole} = I_{CM} + MD^2 = \frac{1}{2}(\frac{M}{4})R^2+(\frac{M}{4})R^2= \frac{3}{16}MR^2
The total moment of inertia of the cam is
I_{total} = I_{disk} - I_{hole} = \frac{35}{48}MR^2 - \frac{3}{16}MR^2 = \frac{13}{24}MR^2
The equation for rotational kinetic energy is
K_R = \frac{1}{2}I\omega^2
So then the kinetic energy of the cam is
K_R = \frac{1}{2}(\frac{13}{24}MR^2)\omega^2 = \frac{13}{48}MR^2\omega^2
However, when I reference my answer with the true answer I am wrong. The true answer is
K_R = \frac{23}{48}MR^2\omega^2
I'm off by \frac{5}{24}. Some one please tell me where I went wrong.