Acceleration of a rolling sphere up a ramp

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Homework Help Overview

The problem involves a bowling ball rolling up a ramp at an angle B, focusing on the forces acting on the ball, its acceleration, and the friction required to prevent slipping. The context is kinematics and dynamics, particularly concerning rolling motion and frictional forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between forces acting on the ball, including gravitational force and friction. There are attempts to derive the acceleration of the center of mass and the conditions for static friction. Some participants question the signs of the acceleration and whether the derived acceleration is consistent with the ball's motion down the ramp.

Discussion Status

The discussion includes various attempts to analyze the problem, with some participants providing calculations and others questioning the correctness of the signs and values. There is an exchange of ideas regarding the acceleration when the ball rolls down the ramp, indicating a productive exploration of the topic.

Contextual Notes

Participants are working under the constraints of treating the ball as a uniform solid sphere and ignoring certain factors like finger holes. There is also a focus on the minimum coefficient of static friction required to prevent slipping, which remains a point of inquiry.

pathmaker
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Homework Statement


"A bowling ball rolls without slipping up a ramp that sloped upward at an angle B to the horizontal. Treat the ball as a uniform, sold spehere, ignoring the finger holes. a) explain why the friction force must be directed uphill. b) what is the acceleration of the center of mass of the ball? c) what minimum coefficient of static friction is needed to prevent slipping?


Homework Equations


I=2/5mr^2
F=ma, torque=fr=I alpha, a=r alpha
f=mu n


The Attempt at a Solution



b) f-mgsinB=ma
fR=Iaplhpa
fR=I(a/r)
f=2/5ma
2/5ma-mgsinB=ma
a=-5/3gsinB

c)f=2/5m(-5/3gsinB)
f= -2/3mgsinB
f/n=mu
mu>-2/3tanB
??
 
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pathmaker said:
b) f-mgsinB=ma
fR=Iaplhpa
fR=I(a/r)
f=2/5ma
2/5ma-mgsinB=ma
a=-5/3gsinB
Careful with signs: The acceleration is down the incline.
 
so a = 5/7gsinB
is that the same as it the ball was moving down the ramp?
 
Yep.
 
thanks
 

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