problem 4 25 wrecking ball of mass m 250kg hanging from fixed point in wire cf length 150m the b

Step 1: Calculate the angle θ using the equation θ = cos^(-1)(1 - v^2 / (2*g*L)), where v = 11.5

Problem 4 25 wrecking ball of mass m 250kg hanging from...

Problem 4: (25%) A wrecking ball of mass m = 250 kg is hanging from a fixed point on a wire of length L = 15.0 m. The ball is released from rest at a position where the wire makes an angle of θ with the vertical. The ball hits the wall just as the wire becomes vertical. The speed of the ball at this instance is v = 11.5 m/s. Use 9.81 m/s^2 for the gravitational acceleration. What was the angle θ? An imaginative engineer suggests using a spring with a spring constant of 180 N/m to shoot the ball against the wall through a horizontal friction guide. Assume that the guide is long enough for the spring to return to its relaxed state before the ball hits the wall. What must be the initial compression of the spring to achieve the same impact speed as in the arrangement with the wire?

Transcript

00:01 Okay, so here we have a question of a wrecking ball which is swinging down and hitting a wall with a velocity of 11 .5 meters per second.

00:10 The wrecking ball is on a cable of length 15 meters.

00:14 It has a massive 250 kilograms and in order to reach the speed, it is pulled up to an angle to the vertical of angle theta and this is what we need to find out.

00:24 Now in order to do this, we're going to need to use the fact that the speed gain.

00:30 Or the kinetic energy gained by the retin ball comes in totally comes entirely from the loss in gravitational potential energy so mgh equals half mv squared this is our equation for gravitational potential energy mass times gravitational field strength times the height gained equals half times the mass times the velocity squared but you can see here that there's no place for angle how do we get the angle included here well, we're going to do that by getting an expression for the height in terms of theta.

01:06 So it's going to be a bit of trigonometry here.

01:09 But the length of the cable is 15 metres.

01:12 That's written there.

01:13 So this is going to be 15 as well.

01:15 Now, you should be able to recognise that that would mean this length here.

01:25 So everything above the green line, that is going to be, that's my regular pen.

01:30 That's going to be 15 cos theta.

01:36 So the height gained is going to be 15, this entire length here, minus what is above the green line.

01:46 Okay, so i'll highlight that there, and highlight that in red.

01:52 So it's going to be this area here.

01:53 That's the height gained.

01:55 And we have an expression for that.

01:57 We can say that delta h is equal to, on this scenario, it's 15 minus 15 cos theta, but as a general formula, it's going to be the length l minus l cos theta, which is l1 minus cos theta, just taking l out of the brackets there.

02:19 So l1 minus cos theta is our expression for the change in height.

02:26 And now we can substitute this into our equation.

02:29 So we have, i'll just write that in again, mg delta h equals half mv squared and actually we can simplify this just very quickly we'll cancel the ms and now i can take this equation for delta h and put it into this equation here so we have g l 1 minus cos theta is equal to half v squared and then i'll divide through the g l over to the other side.

02:57 So 1 minus cos theta is equal to v squared over 2gl and that 2 here just comes from the half and then what i'm going to do is i'm going to make the cos theta positive.

03:10 So i'm going to move the cos theta over to the other side and take away v squared over 2gl move to that side...

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