Physics: friction force: how far will the curling stone glide?

The tutor gives an example with friction force.

I’ve never been curling. Even so, I can imagine the following question resonates with many curlers and spectators:

A curling stone is released at 1.2m/s. If the coefficient of friction between the ice and the stone is 0.010, how far will it travel before coming to rest?

Solution:

From my previous post, the force of friction is calculated by

Ff = μFN

In this case, the stone is on level ice, so

Ff = μFN = μmg

The force of friction is the only unbalanced force acting on the stone, so its acceleration is due only to friction. From Newton’s Second Law,

a = F/m = μmg/m = μg = 0.010(9.8) = 0.098m/s2

Since the acceleration opposes the velocity, its sign is negative:

a = -0.098m/s2

Now, we can calculate the distance the stone travels using kinematics:

v22 = v12 + 2ad

where v1= initial velocity = 1.2m/s, v2 = 0m/s

Therefore,

02=1.22 + 2(-0.098)d

0 = 1.44 – 0.196d

We add 0.196d to both sides:

0.196d = 1.44

We divide both sides by 0.196:

d = 1.44/0.196 = 7.3469…m (7.3m in sig figs)

Apparently the stone will glide 7.3m before stopping.

Source:

Heath, Robert W et al. Fundamentals of Physics. D.C. Heath Canada Ltd., 1981.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physics: projectile motion: maximum height

Continuing about projectiles, the tutor finds the maximum height of the golf ball.

My post from Oct 16 sets the premise: a golf ball is struck at 55m/s at 27° elevation. The velocity is resolved thus:

    \[v_x=49m/s\]

    \[v_y=25m/s\]

To find the maximum height, we focus on v_y. Specifically, when v_y=0, the ball is at max height.

We can use the formula

    \[v_f^2=v_i^2+2ad\]

in which v_f is the final velocity; v_i, the initial; a, the acceleration due to gravity; d, the displacement (in this case, the vertical displacement: the height).

    \[0^2=25^2+2(-9.8)d\]

    \[0=625 -19.6d\]

    \[19.6d=625\]

    \[d=\frac{625}{19.6}=32m\]

The golf ball reaches max height of 32m.

HTH:)

Source:

Giancoli, Douglas C. Physics. New Jersey: Prentice Hall, 1998.

Thanks to quicklatex

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physics: projectile motion: range of a projectile

Continuing with the golfing premise, the tutor finds the range of the projectile.

While this post continues from yesterday’s, its premise begins with my Oct 16 one.

We have a projectile whose horizontal velocity is 49m/s (see Oct 16), whose flight time is 5.2s (from yesterday).

The range of a projectile means how far it travels, horizontally, before landing. The golf ball will bounce and roll afterwards, but to us, that’s not part of the range.

High school physics generally neglects air resistance; therefore, we are as well. Then, the ball’s horizontal velocity remains a constant 49m/s throughout its flight. The range is given by

d=vt=49(5.2)=254.8m=250m (2 significant figures)

Finding the range of a projectile is easy once you know its horizontal velocity and flight time:)

Source:

Giancoli, Douglas C. Physics. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physics: average velocity

The tutor illuminates a very plain, yet critical, concept from physics.

Average velocity has the following definition:

    \[v_{ave}=\frac{displacement}{time}\]

Displacement means your change in position. (Read more about the difference between distance and displacement in my post here.)

Travelling straight-line, displacement and distance are numerically equivalent, although for displacement you need to report the direction.

Let’s consider the situation of straight-line travel, but with velocity that is not consistent:

Example. A motorist travelling East drives for three hours at 80km/h, then stops for a 45 minute lunch, then continues East at 70km/h for another 40 minutes. What is the average velocity?

Solution:

We refer to the formula

    \[v_{ave}=\frac{sum\ of\ displacements}{total\ time}\]

Using the formula

    \[displacement=velocity \times time\]

we can find the driver’s displacement along each leg of the journey, then add them. We divide that sum by the total travel time, including the break. Of course, the minutes need be converted to hours:

    \[v_{ave}=\frac{(80 \times 3 + 0 \times \frac{45}{60} + 70 \times \frac{40}{60})}{(3 + \frac{45}{60} + \frac{40}{60})}\ \frac{km}{h}\ East\]

which simplifies to

    \[v_{ave}=\frac{286.67}{4.42}= 64.9  \frac{km}{h}\ East\]

Due to breaks, and even stopping and starting, a person’s average velocity can be significantly less than their driving speed:)

Source:

Heath, Robert et al, Fundamentals of Physics. D.C. Heath Canada, Ltd, 1981.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.