1) 2) 3) 4) A mass m = 6 kg hangs on the end of a massless rope L = 1.92 m long.
ID: 2134150 • Letter: 1
Question
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A mass m = 6 kg hangs on the end of a massless rope L = 1.92 m long. The pendulum is held horizontal and released from rest. How fast is the mass moving at the bottom of its path What is the magnitude of the tension in the string at the bottom of the path? If the maximum tension the string can take without breaking is Tmax = 487 N, what is the maximum mass that can be used? (Assuming that the mass is still released from the horizontal and swings down to its lowest point.) Now a peg is placed 4/5 of the way down the pendulum's path so that when the mass falls to its vertical position it hits and wraps around the peg. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point). How fast is the mass moving at the top of its new path (directly above the peg)? Using the original mass of m = 6 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)? A mass m = 80 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 16.5 m and finally a flat straight section at the same height as the center of the loop (16.5 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.) What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? m/s What height above the ground must the mass begin to make it around the loop-the-loop? m If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? m/s If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (16.5 m off the ground)? m/s Now a spring with spring constant k = 15300 N/m is used on the final flat surface to stop the mass. How far does the spring compress? m It turns out the engineers designing the loop-the-loop didn't really know physics - when they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning. How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track? m/s The work done by the normal force on the mass (during the initial fall) is: positive zero negative A 5 kg block is placed near the top of a frictionless ramp, which makes an angle of 30o degrees to the horizontal. A distance d = 1.3 m away from the block is an unstretched spring with k = 3 times 103 N/m. The block slides down the ramp and compresses the spring. Find the maximum compression of the spring. xmax
Explanation / Answer
Initially it was at a point O, It is now at a point (3/5*2.17) = 1.302 m
From rest it has gained a speed v .
v^2 = 2gh = 2*9.8*1.302
v = 5.05m/s
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When the bob reaches its bottom most position the string has the high tension
T = mg + mv^2/r = mg + m (2*g*h) /r = mg + m (2*g*h)/ (h/5)
T = mg + m ( 2*g*h)/(h/5) = 11mg
495 =11*9.8 m
m = 4.59 kg
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Additional information
The peg position affects the tension in the string.
If there were not a peg, the centripetal force will be mv^2/L
But since there is a peg, the radius of the circular path is reduced at this instant and hence the centripetal force needed is more than mv^2/L it is now increased to 5 (mv^2/ L)
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