- #1
Joshua Benabou
<Moderator's note: Moved from a technical forum and thus no template.>
We place a ball on an an inclined plane (angle ##\theta##). At the top of the ball is a small dog who always stays at the top over the course of the ball's movement.
What is the motion of the ball?
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I apologize for the lack of diagram.
Let's note:
- mass of ball=##M##
- mass of dog=##m##
- point ##Q## =center of ball, point ##P##=position of dog at the top of the ball
- radius of ball=##R##
- moment of inertia of ball = ##I=kMR^2##
- ##f_d## =static friction force applied horizontally at ##P## to the ball by the dog
- ##f_b##=static friction force applied parallel to the inclined plane to the ball by the plane
- ##N_d## the normal force between the dog and the ball
- ##N_b## the normal force between the ball and the inclined plane
- ##\alpha##=magntitude of angular acceleration of the ball
- ##a##=magntitude of acceleration of center of mass of the ball
We are looking for ##\alpha## and ##a##.
We have the following equations:
1. ##I\alpha=(f_d+f_b)R## (torque applied to the ball)
2. ##Ma=f_d\cos\theta+N_d\sin\theta-f_b+Mg\sin\theta## (projection of forces applied to ball in the direction parallel to the incline)
3. ##mg\sin\theta-f_d\cos\theta-N_d\sin\theta=ma## (since the vector ##QP## is constant, the dog has the same linear acceleration as the ball; this equation comes from projecting the forces applied to the dog in the direction parallel to the incline)
4. Condition for rolling without slipping: ##a=R\alpha##
Note that projecting forces along the direction perpindicular to the incline will give use 2 more equation (one for the forces applied to the ball, one for the forces applied to the dog), but they won't be useful.
The above equations are enough to determine ##a## in terms of the force ##f_d##, but that's the best we can do.
Is it thus necessary to make a choice about how to model the dog? What conditions am I missing?
We place a ball on an an inclined plane (angle ##\theta##). At the top of the ball is a small dog who always stays at the top over the course of the ball's movement.
What is the motion of the ball?
------------------
I apologize for the lack of diagram.
Let's note:
- mass of ball=##M##
- mass of dog=##m##
- point ##Q## =center of ball, point ##P##=position of dog at the top of the ball
- radius of ball=##R##
- moment of inertia of ball = ##I=kMR^2##
- ##f_d## =static friction force applied horizontally at ##P## to the ball by the dog
- ##f_b##=static friction force applied parallel to the inclined plane to the ball by the plane
- ##N_d## the normal force between the dog and the ball
- ##N_b## the normal force between the ball and the inclined plane
- ##\alpha##=magntitude of angular acceleration of the ball
- ##a##=magntitude of acceleration of center of mass of the ball
We are looking for ##\alpha## and ##a##.
We have the following equations:
1. ##I\alpha=(f_d+f_b)R## (torque applied to the ball)
2. ##Ma=f_d\cos\theta+N_d\sin\theta-f_b+Mg\sin\theta## (projection of forces applied to ball in the direction parallel to the incline)
3. ##mg\sin\theta-f_d\cos\theta-N_d\sin\theta=ma## (since the vector ##QP## is constant, the dog has the same linear acceleration as the ball; this equation comes from projecting the forces applied to the dog in the direction parallel to the incline)
4. Condition for rolling without slipping: ##a=R\alpha##
Note that projecting forces along the direction perpindicular to the incline will give use 2 more equation (one for the forces applied to the ball, one for the forces applied to the dog), but they won't be useful.
The above equations are enough to determine ##a## in terms of the force ##f_d##, but that's the best we can do.
Is it thus necessary to make a choice about how to model the dog? What conditions am I missing?
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