## Dynamics of Circular Motion Problem

Source: Past exam (Kuwait University) 2nd Midterm, Fall Semester 2015

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A pilot of mass m=60 kg in a jet aircraft executes a loop-the-loop, as shown in the figure. The aircraft moves in a vertical circle of radius 800 m at a constant speed of 200 m/s. Determine the force (in N) exerted by the seat on the pilot at the top of the loop.

Let’s analyze the paragraph above:

1. Aircraft doing a vertical circular loop “…aircraft executes a loop-the loop,…. moves in a vertical circle…”. Then the problem is a circular motion problem. Knowledge about radial acceleration should be present.
2. “…The force exerted by the seat on the pilot…” means it is a normal force. Normal force commonly does not present alone.. we should think about other forces that are present as well.

What are the given information:

1. Pilot mass (m) = 60 kg
2. The circular loop radius (R) = 800 m
3. The speed of the plane (v) = 200 m/s
4. Tangential acceleration (at)=0 m/s2 (clue: the two words “constant speed”).

What the problem need:

1. The normal force on the pilot at the top (shown in figure above)

SOLUTION:

The pilot diagram at top of the loop:

The normal force (n) directed from seat to pilot, hence when the aircraft at top of loop and pilot is up side down the normal force is directed downward.

The pilot weight (Fw) directed from pilot toward the ground, hence it is directed downward.

From the below image, with a constant speed, the total acceleration rises from the circular motion is the radial acceleration (ar).

## Projectile Problem

From Young and Freedman’s University Physics with modern physics, 14th edition

Lets analyze the wording of the problem:

1. A coin will be thrown into the air to fall into a dish “…you toss a quarter into a dish…”.. then it is a projectile problem.
2. The coin should land into a dish that is in a higher height than were the coin tossed is. We can take the height of the tossed coin as y1=0 m (because the height of dish needed is from where the coin left the hand) and the height of the dish as y2=h.

What are the given information:

1. initial velocity (Vo) = 6.4 m/s.
2. Angle of projected coin = 60°
3. Δx=2.1 m
4. The height where the coin (y1) assumed to be 0 m

What the problem need:

1. The height where the dish is (y2=h).
2. The vertical component of the final velocity.

SOLUTION:

1. The projected coin path could be seen in the below diagram:

## Relative velocity problem

A problem from past exam of physics 1 (Kuwait University)
First midterm, First semester (Fall semester) 2001

Two ships A and B, leave port at the same time. Ship A travels due east with a speed of 20 km/h with respect to the earth and ship B travels 48 degrees north of east with a speed of 30 km/h with respect to the earth. Determine the velocity of A relative to B.

lets analyze the wording of the problem:

1. By looking at  ” velocity of A relative to B” and “Ship A travels …. with respect to the earth” … then it is a relative velocity problem

The given information in problem:

1. Ship A travels to east with speed 20 km/h relative to ground (earth).
2. Ship B travels to the direction of north of east by a 48° with a velocity of 30 km/h relative to the ground (earth).

What the problem need:

1. The velocity of A relative to B

SOLUTION:

The diagram below shows the directions of ship A and ship B.

Ship A velocity is 20 km/h in the direction of positive x-axis with an angle zero degree with the x-axis that is why it is shown as a horizontal velocity. The only component of the velocity is in the direction of positive x-axis.

Ship B has a direction that is 48° from the positive x-axis, hence this ship’s velocity has a component in the x-axis and a component in the y-axis.

## Circular motion with constant acceleration problem

This problem from Serway and Jewett’s Physics for scientists and engineers with modern physics textbook (9th edition)

Let’s analyze the wording of this problem:

1. A ball swings in a vertical circle… then it is a circular motion problem.
2. A given acceleration at specific position in space which is 36.9° past the lowest point on way up for the ball..then the given acceleration does not seem not to be a radial acceleration (see the diagram below in solution).

The information given in this problem:

1. Radius of circular motion is 1.5 m (R=1.5m).
2. The acceleration of the ball is a=-22.5iˆ+20.2jˆ m/s² at 36.9° away from the vertical axis (y-axis).

What the problem need:

1. Sketch the vector diagram showing the components of the acceleration.
2. Find the magnitude of radial acceleration.
3. Find the speed and velocity of the ball.

SOLUTION:

1. Sketch the vector diagram showing the components of the acceleration.

The diagram of acceleration vector

2. Find the magnitude of radial acceleration.

The magnitude of radial acceleration could be found by using the horizontal and vertical components of the given acceleration:

3. Find the speed and velocity of the ball.

The speed could be found using the radial acceleration and the radius of the circular motion:

## Projectile Problem

Problem from Young and Freedman’s University Physics with Modern Physics textbook

First we need to analyze the wording of the problem:

1. A snow ball rolls off a barn roof.. that mean it is a projectile problem
2. The height is y1=14 m and y2=0 m (the ground) then Δy=-14 m
3. The range is x2= 4 m and x1=0 m (the barn side has the x-axis origin).

What are the given information:

1. Angle at which the ball is leaving the roof is 40°.
2. The initial velocity (vo) =7 m/s
3. Δy=-14 m
4. Δx=4 m

What the problem need:

1. How far the ball will fall away from the barn (Δx=?)
2. Draw x-t, y-t, vx-t, and vy-t graphs for the motion in part (1)
3. If a man (his height= 1.9m) is standing at Δx=4 m, do the ball hit him when it falls?

SOLUTION:

1. How far the ball will fall away from the barn (Δx=?)

2. Draw x-t, y-t, vx-t, and vy-t graphs for the motion in part (1)

3. If a man (his height= 1.9m) is standing at Δx=4 m, do the ball hit him when it falls?

For this question, it is easy to find the time when Δx=4 m:

While the man height is 1.9 m then the ball will be far from hitting him and he will be fine.