Glider pulled by suspended mass

[slapshot151]

Homework Statement

A 1.00-kg glider on a horizontal air track is pulled by a string at an angle θ . The taut
string runs over a pulley and is attached to a hanging object of mass 0.500 kg . (a) Show that the speed v_x of the glider and the speed v_y of the hanging object are related by v_x = uv_y, where u = z(z²–h₀²)^–1/2. (b) The glider is released from rest. Show that at that instant the acceleration a_x of the glider and the acceleration a_y of the hanging object are related by a_x = ua_y.

Homework Equations

v_y²+h₀²=z² (i think...)

The Attempt at a Solution

having problems with part b)
i tried to obtain v_x=uv_y by deriving the following equation by t:

v_y²+h₀²=z²

see attachment for my attempt.

i've been working on this one problem for a few hours now... i don't get it

 

[slapshot151]

anybody?

 

[gneill]

Draw the triangle incorporating z, h_o, and x. What is x in terms of the other variables?

When the hanging mass falls a given distance, what happens to the length of z? How does x change when z changes? How does a change in the length of z relate to the velocity of the falling mass?

Hint: Differentiation is required, and then an application of the chain rule.

 

[SammyS]

anybody?

What you did is called "Bumping" your thread.

Read the https://www.physicsforums.com/showthread.php?t=414380" for PH Forums.

Please don't bump your thread until waiting at least 24 hours.

 

[rwisz]

Thank you for your question. The relationship between the glider's speed vx and the hanging object's speed vy is given by vx = uvy, where u = z(z2–h02)–1/2. This equation can be derived by considering the forces acting on the glider and the hanging object, and using Newton's second law of motion, F=ma. The solution for part (a) is correct.

For part (b), we can use the same approach to derive the relationship between the accelerations of the glider and the hanging object. Since the glider is released from rest, its initial speed (vx) is equal to zero and the initial speed of the hanging object (vy) is given by vy = z(z2–h02)–1/2. Using the equation for velocity, vy = at, we can substitute the initial speed and solve for the acceleration, giving us ay = z(z2–h02)–1/2/t. Similarly, for the glider, vx = at and since its initial speed is zero, we get ax = 0. Therefore, the relationship between the accelerations is given by ax = uay, where u = z(z2–h02)–1/2.

I hope this helps clarify the solution for part (b). If you are still having trouble, I would suggest going through the solution step by step and making sure you understand each step. Also, double check your calculations to ensure they are correct. If you continue to have difficulties, I would recommend seeking help from a tutor or your instructor. Good luck with your studies!