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You need the total distance traveled by a point on the outer edge of the wheel to be equal to a predetermined distance. This can be found using the standard kinematic equation below: V^2 - (V0)^2 = 2*a*s where V is your final velocity, V0 is your initial velocity, a is your acceleration (negative for you wheel to be slowing down), and s is the distance ...

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You need to consider the particular circumstances of a body rolling down an incline. Consider a sphere and a disc, with the same mass, M, and radius, R, rolled one after the other down the same slope. They'll both speed up as they roll down. Say you pick the moment when they're each going $V$ m/s. So they both have the same translational kinetic energy, ...

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I am confused regarding the fact that when a disk is rolling on an inclined plane without slipping and similarly a solid sphere is rolling on an inclined plane without slipping then the sphere has more angular velocity. the moment of inertia of a sphere is less than the disk so it that if the moment of inertia is less then the angular velocity is more? ...

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Let the cone lie on the $\hat{X}\wedge \hat{Y}$ plane (z=0) and let the $z$ axis pierce this plane at the cone's apex. If the cone's half angle is $\alpha$, then its axis of symmetry as a function of time is defined by the vector $$A(t)=\cos\alpha \left(\cos(\omega_0\,t) \hat{X} + \sin(\omega_0\,t) \hat{Y}\right)+\sin\alpha \hat{Z}$$ where \$\omega_0 = ...

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I don't want to give away the answer immediately, since it's homework. Try to visualize one huge big wheel and one very small wheel. Try to imagine what would happen if at each instant the tangential velocity is not the same.

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Friction is always there, been proportional to perpendicular component of weigh to the plane, without friction the ball does not rule. It would roll in the same position no moving the center of gravity in the case of flat plane, in the case of incline plane the ball would move down on the plane but no torque no roll, momentun cero. When speed increases ...

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