Trajectory of Mud Glob Thrown From Wheel: Solving the Mystery

[physicsnoob1]

Homework Statement

A wheel of radius r is rolling along a muddy road with speed v, and particles of mud are being continuously thrown off from all points of the wheel. ignoring air resistance, what is the trajectory of a mud glob thrown off the wheel when it is at angle (theta) to the horizontal? at what (theta) is the height maximized? What is the meaning of the critical speed v^2 = rg?

Homework Equations

v = 2(pi)r / T is the propagation velocity of the wheel. but my first question is, is this the same as the rotational velocity? I've thought about it for a while and it seems like it is. i am stumped as to how to find the trajectory of the mud glob though. please help!

The Attempt at a Solution

well i guess I am still stuck on figuring out if the propagation velocity is the same as the rotational velocity. after that i was thinking about playing with the velocity vectors to find an expression for the trajectory. the only force acting on the glob should be gravity i think.

 

[Hootenanny]

Welcome to Physics Forums.

Have you tried to sketch the problem? You know that when the wheel is at some angle, theta, a mud particle flies off. In what direction relative to the wheel will the particle travel?

 

[Richyfeller]

Homework Statement

A wheel of radius r is rolling along a muddy road with speed v, and particles of mud are being continuously thrown off from all points of the wheel. ignoring air resistance, what is the trajectory of a mud glob thrown off the wheel when it is at angle (theta) to the horizontal? at what (theta) is the height maximized? What is the meaning of the critical speed v^2 = rg?

Homework Equations

v = 2(pi)r / T is the propagation velocity of the wheel. but my first question is, is this the same as the rotational velocity? I've thought about it for a while and it seems like it is. i am stumped as to how to find the trajectory of the mud glob though. please help!

The Attempt at a Solution

well i guess I am still stuck on figuring out if the propagation velocity is the same as the rotational velocity. after that i was thinking about playing with the velocity vectors to find an expression for the trajectory. the only force acting on the glob should be gravity i think.

Please help me to show that, if v^2=gr, no mud can be thrown higher than r+v^2/2g+gr^2/2v^2 above the ground,

 

[Richyfeller]

Please help me to show that, if v^2=gr, no mud can be thrown higher than r+v^2/2g+gr^2/2v^2 above the ground,

Welcome to Physics Forums.

Have you tried to sketch the problem? You know that when the wheel is at some angle, theta, a mud particle flies off. In what direction relative to the wheel will the particle travel?

yes i sketch the problem, but the equations to use for the proof is my problem

 

[rwisz]

I would approach this problem by first clarifying that the propagation velocity and rotational velocity are indeed the same thing. This can be confirmed by considering the motion of a single point on the wheel - it will have both a linear velocity (propagation velocity) and an angular velocity (rotational velocity) that are equal in magnitude.

Next, I would consider the motion of the mud glob as a projectile, with an initial velocity that is a combination of the wheel's linear velocity and the tangential velocity of the point on the wheel where the glob was thrown. This can be represented by a vector diagram.

To find the trajectory of the mud glob, I would use the equations of projectile motion, taking into account the initial velocity, angle of projection (theta), and the effects of gravity. The trajectory will be a parabola, with the maximum height occurring when the projectile reaches its highest point. To find the angle at which the height is maximized, I would use the equation for the maximum height of a projectile and solve for theta.

The critical speed v^2 = rg refers to the speed at which the centrifugal force (caused by the rotational motion of the wheel) is equal to the force of gravity. This means that at this speed, the mud glob will continue to travel in a straight line instead of being thrown off the wheel. This is an important concept in understanding the dynamics of rotating objects.

In summary, the trajectory of the mud glob can be found by considering it as a projectile and using the equations of projectile motion. The critical speed v^2 = rg is the speed at which the centrifugal force is equal to the force of gravity and has implications for the motion of the mud glob.