Direction of Friction in Circular Motion: Solving the Record Spin Dilemma

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In a spinning record scenario, the force of friction acts inward, providing the necessary centripetal force to keep an object moving in a circular path. Although the object remains stationary relative to the record, it experiences centripetal acceleration due to its circular motion. The friction between the object and the record is essential for maintaining this motion, as it prevents the object from sliding off. The reaction force from the object on the record is negligible and does not affect the record's motion. Thus, in this context, friction equates to centripetal force, ensuring the object stays in its circular trajectory.
ndogg
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If an object is placed on a record that is spinning, what is the direction of the force of friction?
 
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Which way is he object accelerating?
Got it now?
 
The object itself is not accelerating because it remains still on the record. So is there 0 friction?
 
remember that acceleration is change in velocity per unit time not speed. Velocity has direction involved! This is centripetal acceleration, so the direction of friction is directly inward. Think of riding in a car around a sharp curve.
 
So the direction of friction is in the same direction as the centripetal force (which is the center of the record)? Shouldn't it be in the opposite direction because there has to be an equal and opposite reaction?
 
Good question. The object sitting on the record is moving in a circle. Now pretend that the record does not exist, but instead, some arbitrary force holds the object in a circular path. You know that this force must be the centripetal force because if it were not, the object would go off on a path tangent to the circle. This arbitrary force is the friction of the record. Now the reaction force is actually the friction from the object acting on the record, but this force is not great enough to alter the record and is therefore negligible.
 
So, in this case, friction = centripetal force?
 
Yes, good job. Just so you know, if the record were actually accelerating with speed too, you would just have two components of acceleration perpendicular to each other, so you can use A squared plus B sq = Csq
 
Thanks! Problem solved.
 
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