Quiz March 2018 (GES 1)

As the cube and the sphere have the same mass and are made from the same material, we can assume them to be of the same volume. Also, we have no dimensions of the cube, so we must start with the sphere. We can do the calculation in mm or in cm, this time I have done it in mm.

now that we know the volume required, let's find the length of the rod

we can use the formula for the volume of a cylinder. We KNOW the volume, and we can transpose to find for h

Let's write out the question,  we can then collect like terms:

Let's cancel what we can, and then rewrite it to make it neater:

Multiply both sides by 4y to move it up, and then take the fifth root of both sides:

and we are done!

Before we get too involved, lets see if we can simplify any other this:

let's rewrite it so we can see it better, and then cancel out what we can

One last step! We need to transpose the equation to make x the subject! (this is quite unusual!)

done!

First let's draw a diagram and write down the formula for the volume of a cone

The question is asking for the volume in meters cubed. It's best we start that way, so let's convert these measurements to meters. We will also need the base, so we can use Pythagoras to calculate the base:

so now we have the height of the cone is 0.3m and the radius is 0.16m, we can calculate the volume in meters cubed directly.

We need to find the angle from the base. We have everything we need, and we can keep the measurements in the original mm to avoid typos:

Let's use SOH CAH TOA to solve for the missing angle

This is a simple question. They are usually hidden towards the end of 2 or 3 parters, so don't give up! We just apply a simple formula for the curved surface area:

For this question, we need to plug in values of x into the equation and calculate a value for y. we can show this all in a neat table. I included a screenshot of my calculator just to show you how to enter it all in, you can backspace and reuse the same equation so you don't have to retype the whole thing each time!

We need to make a neat graph, you will get points for

• y-axis vertical, x-axis horizontal
• equal spacing on the x and y axis
• clear points marked
• a smooth hand-drawn line for the parabola

We can use the graph to find an estimate for the value of x when y=5. We just need to draw a horizontal line at y=5. Then we can see where it touches the curve, and trace these points down to the x-axis, this gives us our two solutions, both are valid.

This is a simple suvat equation to find the acceleration, from there we will use F=ma to find the force required to create the acceleration. Let's start with a diagram and stating what we know:

We can use the second suvat equation stated to find the acceleration by transposing.

Now we know the acceleration (we know the mass is 100kg) we can calculate the force required using F=ma:

This is the second part of this question, so we need to model it as a new set of SUVAT equations. We have the initial velocity as 3.5 m/s and the final as zero. we also know the acceleration, remember to use a negative 0.5 as it is retardation.

Using the same SUVAT equations as the previous part, we can use the equation v =u +at

a simple transpose and we can solve for the time. Now we have 7 seconds as the time taken to come to rest.

It is very common to have to solve for the two unknowns in SUVAT equations, so after we have found the time, we will have to find the distance traveled, this can be done using the other SUVAT equation:

in this example my student requested I use D for distance rather than S, so I did! We are dealing with a lot of formulas, so I usually encode these as mnemonics to help us remember them. The mnemoic may be different for each person who learns it, also I work with a lot of students with dislexia, so sometimes it helps to change the letters. Whatever works!

This is a simple plug in and solve:

To calculate the linear velocity we need to find the distance traveled every second by a point on the rim. We will be using the simple formula for speed (its basically the same as velocity, and will make it easier to remember). So we need the distance traveled every second.

The distance traveled by a point on the rim will be the circumference of the circular path traveled, so we need the circumference, which we just calculated above. This is the distance traveled in one rotation.

Next, we need to see how many rotations the flywheel is doing per min, the question is a little harder as they gave us rads per min, so let's convert into revolutions (rotations)

In order to get the distance traveled, we multiply the circumference by the number of revolutions, and we can write this over 1 min as a fraction as we were given the number of rads per min.

the last step here is to convert the distance traveled in 60 seconds down to 1 second, giving us 67.29 m/s as the answer.

we calculated this in the previous section!

We need to convert from degrees to radians. This is very common in this section of the test, to be able to convert between radians, degrees and revolutions. We will make a table, starting with a fact we know: 360º= 2pi radians.

Then we will convert 360 into the required 150. Todo this we will divide 360 by 360 to get 1, then multiply by 150 to get 150 (just like we do with percentages!) anything we do to one side, we will also to do the other, this will maintain the proportional relationship. Again we have to ALOT of conversions as an engineer, so building this table allows for easy conversion of units.

There is a problem with this question. It doesn't say if the force is an elevated force (going up) or a depressive force (pushing down). We should assume that it's going up (always) as otherwise, the box will not move for part b!

First we will draw a diagram, then work on splitting that diagonal force into its Vertical and horizontal components. We will use SOH CAH TOA to split up the force. This is important because the HORIZONTAL force will be pushing (or pulling its the same) the box along the plane, and the VERTICAL force will be contributing to, or reducing the friction felt by the box.

So now we have the horizontal push, we can calculate the acceleration, the vertical "lift" doesn't affect this part of the question, as we have no friction, but we will need it for part b.

First, we need to calculate how much the box is pushing down on the ground. As its being pulled down by gravity, we use F=mg (same as F=ma) to calculate the downwards force as 392.4N.

Next, we need to find the net vertical force, so we are pulling the box upwards with our applied force, so we need to subtract this:

using this vertical force, we can calculate the total amount of friction force felt on the box:

So this is the resisting force. We now need the net horizontal force, so we can take our pushing force, and subtract this frictional force. The force that is left over we will use F=ma to calculate the new acceleration:

This question is a quite a common problem about geometrical solids. We can break the shape into a few sections. There is a hemisphere cap, and a cylindrical body, but I will go further and break it into the SWEPT volume and the CLEARANCE volume.

The question explains that the clearance volume is 95% of the cylinder, so lets find that first:

So now we can work on finding the clearance volume, this is made up of two sections, the hemisphere above and a small 4cm cylinder:

Now lets work on the SWEPT volume, its a simple cylinder

to find "the clearance volume as a percentage of the swept volume" we will need to divide one by the other and then multiply by 100 to turn into a percentage: