As a new engineer for the ABC Sports Equipment Company, you have been given the

Question

As a new engineer for the ABC Sports Equipment Company, you have been given the task of specifying design parameters for several components of the "Super Duper" Outfielder Trainer. This device launches baseballs to an outfielder from home plate. It uses a simple spring-block contact design. With this design a block which is driven by a spring makes the necessary contact with baseball to launch it into flight to the outfielder.
The weight of the block is 5-lb, the mass of a typical baseball is 0.146 kg (that is, 5.125 oz), and the coefficient of restitution is approximately 55%.

1.    You are required to determine the minimum value of the spring constant (k) necessary to achieve a
horizontal distance of 400 ft if the range of is between 15 to 75. The maximum distance of compression is 1.5 ft, and at the point of block ball contact the spring is in the undeformed position. Wind and air resistance should be neglected. However, the friction coefficient between the block and the rod, which supports it, is 0.05.

2.    Plot, at least in table format your results for k for the given range of in 5 increments. 3.    For = 60, it is determined that the average time of contact between the block and the ball is 0.005 second. Assuming a constant impulsive force is applied during that time interval, calculate the
magnitude of (a) the impulse and (b) the force.
Note: Use SI Units in your computations.

Explanation / Answer

1. mass of ball = 0.146 kg
mass of block = 5lb =0.45 x 5 = 2.25 kg
Now, for horizontal distance = 400ft = 400 x0.305 m = 122 m
Now, maximum value of range is only when, = 45o

In this case, u2 sin(90o)/g = u2 / 9.8 = 122 , So, u = 34.58 m/s.

Now , since coefficient of restitution =55% , So , speed of block = 34.58 x 1/0.55 = 62.87 m/s

Now, maximum compression = 1.5 ft = 0.4575 m

now, total energy in the block till the release of 5 lb block = 1/2 m v2 = 1/2 kx2 - mgx(sin 45) - mgsin(45)

or, v2 = (k/m)(0.4575)2 - 2 x 9.8 x (1/2) -2 x 0.05 x 9.8 x(1/2) = 62.872

So, (k/2.25) = 18920

or, k = 42572 N/m

I am unable to do question number 2 and 3. It will take very long time, + i am unable to understand the 2nd part question. question number 3 is based on 2nd question.

really sorrry friend, i did as much i could.

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