Will the Cake Stay on the Car in the Rotary?

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A driver traveling at 10 m/s in a rotary with a 15 m radius places a cake on top of the car. The friction coefficient between the cake and the car is 0.600. To determine if the cake will stay on, one must analyze the forces acting in both the x and y directions and calculate the required centripetal acceleration. The maximum frictional force must be compared to the centripetal force needed; if the friction is insufficient, the cake will slip off. In this scenario, a coefficient of friction of at least 0.68 is necessary to keep the cake from falling.
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A driver leaves a cake on top of his car and enters a rotary with a speed of 10 m/s. The rotary has a radius of 15 m and the coefficient of friction between the cake and the top of the car is 0.600u. Will the cake stay on top?

Can someone show both ways of doing this problem?
 
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Draw a free body diagram. What forces are acting in the x and y directions? How does the force in the x-direction relate to centripetal motion?
 
This isn't a homoework problem fwiw, it was on a test. I solved it but wanted to know if I missed something and or the other way to do it.

Fnet=ma find acceleration then find acceleration of car in rotary. Ac in rotary > max a of cake. The cake then falls off.
 
The friction force between cake and roof will provide the centripetal acceleration necessary to keep the cake moving in a circle. Find the maximum frictional force, and if this is less than the centripetal force needed, then the cake will slip. I believe that in this case the coefficient would need to be at least 0.68
 

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