Chapter 2: Newtonian theories of motion

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Last updated 2:14 PM on 1/4/26
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23 Terms

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What is Newton’s first law?

Every object continues to be at rest or continues with a constant velocity unless it experiences an unbalanced force.

Not observed in everyday life due to friction and air resistance which eventually slow the motion of the object. Can be balanced with some other forces.

For example” constant velocity can be maintained if driven by a motor.

If force exerted = force applied, object remains stationary.

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What is Newton’s second law?

An object will accelerate at a greater rate when the force acting on it increases, and heavy objects harder to accelerate than lighter ones.

Fnet=ma

Fnet is net force

m=mass of object

a=acceleration of object (ms-2)

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What is Newton’s third law?

When a body exerts force on another body, the second body exerts an equal force on the first body but in the opposite direction.

While the force is the same size on both objects, the resulting acceleration may not be. For example, stubbing toe on heavy rock causes toe to decelerate significantly, but rock doesn’t move due to much greater mass (2nd law).

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What is speed relative to a circle?

speed=d/t=circumference/period=2πr/T

Imagine an object moving in a circular path of radius r metres with a constant speed of v and takes T seconds to complete one revolution. The time taken to travel once around a circle is called the period, T, of the motion.The number of rotations each second is the frequency,f
f=1/T

<p>speed=d/t=circumference/period=2πr/T</p><p></p><p>Imagine an object moving in a circular path of radius r metres with a <strong>constant </strong>speed of v and takes T seconds to complete one revolution. The time taken to travel once around a circle is called the period, T, of the motion.The number of rotations each second is the frequency,f<br>f=1/T</p>
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What is centripetal acceleration?

When an object moves in a circle, its speed can stay constant, but its direction is always changing.
That change in direction requires an acceleration — this is centripetal acceleration.

a=v²/r, where v=2πr/T

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What is centripetal force?

Centripetal force is the net inward force that causes an object to move in a circular path. For example, when considering a hammer throw, the centripetal force would be the the tension force in the cable. Once that is released, there is no longer an inward force causing the hammer to change direction, the result is that the hammer moves in a straight line tangential to its circular path.

Fnet=mv²/r

Fnet is the centripetal force on the object

<p>Centripetal force is the net inward force that causes an object to move in a circular path. For example, when considering a hammer throw, the centripetal force would be the the tension force in the cable. Once that is released, there is no longer an inward force causing the hammer to change direction, the result is that the hammer moves in a straight line tangential to its circular path.</p><p>F<sub>net</sub>=mv²/r</p><p>F<sub>net</sub> is the centripetal force on the object</p>
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What are banked tracks? Benefits?

Consider a road, a horizontal road travelling in a circle relies on the friction between the tires and road to keep it in its circular path. This is the sideways force. A banked track involves angling the road to reduce the need for a sideways frictional force, allowing the car to travel faster without skidding off the road.

The banking angle can be found using the relation tan 0 =Fnet/Fg, remembering Fnet=mv²/r and Fg=mg. Hence, tan0=v²/rg

<p>Consider a road, a horizontal road travelling in a circle relies on the friction between the tires and road to keep it in its circular path. This is the sideways force. A banked track involves angling the road to reduce the need for a sideways frictional force, allowing the car to travel faster without skidding off the road.</p><p></p><p>The banking angle can be found using the relation tan 0 =Fnet/Fg, remembering Fnet=mv²/r and Fg=mg. Hence, tan0=v²/rg</p>
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<p>Calculate the net force acting on the cyclist at this instant and calculate the design speed for this section of the track.</p>

Calculate the net force acting on the cyclist at this instant and calculate the design speed for this section of the track.

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Research why angle between FN and Fnet cannot be theta.

<p>Research why angle between FN and Fnet cannot be theta.</p>
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<p>Calculate the normal force from the track at point Y. Likewise for point Z.</p>

Calculate the normal force from the track at point Y. Likewise for point Z.

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vy=4.85ms-1

FNy=15.58N up

vz=3.704ms-1

FNz=9.701N down

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<p>Calculate the time the ball takes to land<br><br>Calculate the velocity of the ball as it lands.<br><br>Calculate the velocity of the ball as it lands.</p>

Calculate the time the ball takes to land

Calculate the velocity of the ball as it lands.

Calculate the velocity of the ball as it lands.

Horizontal is vav=s/t

Vertical is

v=u+at
s=ut+1/2at²

v²=u²+2as

s: Displacement (change in position) of the object, typically measured in meters (m).

u: Initial velocity of the object, measured in meters per second (m/s).

t: Time duration over which the motion occurs, measured in seconds (s).

a: Constant acceleration of the object, measured in meters per second squared (m/s²)

<p>Horizontal is v<sup>av</sup>=s/t</p><p>Vertical is</p><p>v=u+at<br>s=ut+1/2at²</p><p>v²=u²+2as<br><br><strong><span>s</span></strong><span>: Displacement (change in position) of the object, typically measured in meters (m).</span></p><p><strong><span>u</span></strong><span>: Initial velocity of the object, measured in meters per second (m/s).</span></p><p><strong><span>t</span></strong><span>: Time duration over which the motion occurs, measured in seconds (s).</span></p><p><strong><span>a</span></strong><span>: Constant acceleration of the object, measured in meters per second squared (m/s²)</span></p><p></p>
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2.47 secs 49.9m

31.4ms-1 at 50.44º below horizontal.

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<p>a What is the athlete's velocity at the highest point in the jump?</p><p>b What is the maximum height gained by the athlete's centre of mass during</p><p>the jump?</p><p>c Assuming a return to the original height, what is the total time the athlete is</p><p>in the air?</p>

a What is the athlete's velocity at the highest point in the jump?

b What is the maximum height gained by the athlete's centre of mass during

the jump?

c Assuming a return to the original height, what is the total time the athlete is

in the air?

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6.11ms-1

0.113m

0.23s