HINTS FOR PROBLEM SET #3

Problem Set #3 – Problem 1

Part a)

You want to find when when the acceleration vector is perpendicular to the position vector.

Two vectors are perpendicular when their dot product is zero.

First find the acceleration vector of the particle by differentiating the position vector twice.  (The first differentiation will give the velocity vector.)

Dot this acceleration vector with the position vector and set equal to zero.

Solving for t gives the instants when the acceleration vector is perpendicular to the position vector.

Part b)

Two vectors are parallel when their cross product is zero.

Cross the acceleration vector with the position vector and set equal to zero. 

Solving for t gives the instants when the acceleration vector is parallel to the position vector.

Problem Set #3 – Problem 2

You are asked to determine a range of values for a variable.

Whenever you are asked to determine a range for a variable, you need to find the minimum and the maximum values allowed for the specific situation.

In this case, you need to find the initial velocity, v₀, that will cause the projectile to reach point B.

This gives the minimum value of v₀.

Then you need to find the initial velocity, v₀, that will cause the projectile to reach point C.

This gives the maximum value of v₀.

Consider an x-y coordinate system where

• the y axis is vertical and positive upward,

• the x axis is horizontal and positive to the right,

• the origin is the point on the ground directly below point A.

You need to find an equation that gives the initial velocity, v₀, in terms of these x and y coordinates. Then you can get the range of values for the velocity. 

You can use the equations for projectile motion.

Obtain the equation for x in terms of t (time) and v₀:

             x = x₀ + v_x0t                         note: v_x0 = v₀

Next obtain the equation for y in terms of t (time) and v₀:

             y = y₀ + v_y0t - 1/2 g t²             note: v_y0 = 0

These are parametric equations, i.e. they are equations in the same variable, t. 

You can eliminate time and reduce these 2 equations in 3 unknowns (x, y and t) to 1 equation in 2 unknowns (x and y).

You will have an equation in x and y that also has v₀ in it. 

Solve this for v₀. 

Substitute the coordinates of B to get the minimum value for v₀.

The coordinates of point B are (7, 0.67)

Substitute the coordinates of C to get the maximum value for v₀.

The coordinates of point C are (12.33, 2)

Problem Set #3 – Problem 3

Part a)

You want to get the largest value of the angle alpha for which the puck will enter the net.

You need to find the value of alpha for which the puck will hit point C. 

Consider an x-y coordinate system where

• the y axis is vertical and positive upward,

• the x axis is horizontal and positive to the right,

• the origin is the point A on the ground.

Thus, the coordinates of point C are (4.8, 1.22). 

You need an equation that gives alpha in terms of these coordinates x and y.

You can get the equation for alpha in terms of x and y in the same way as the previous problem. 

First get equations for x and y in terms of t (time). 

Then eliminate time.

You will have an equation with x, y, v₀ (=170 km/h) and alpha.

Substitute the coordinates of C for x and y and then solve this for alpha.

Part b)

To get the corresponding time, just go back to the equation for x in terms of t.

Substitute the value you found for alpha in part a) and x = 4.8 into this equation and solve for t.

This is the time when the puck reaches C.

Problem Set #3 – Problem 4

This is a problem dealing with relative velocities. 

Part a)

You want to find the velocity of the belt so that the velocity of a piece of coal, C, with respect to the belt, B, is vertical when it hits the belt. 

That means that the x component of the relative velocity,  v_C/B,  must be zero.

Use the equations of projectile motion to find the absolute velocity of C as it hits the belt, v_C.

               v_C = (-1.157 i + -5.5974 j)

Find the expression for the absolute velocity of the belt, v_B.

               v_B = v_B(-cos10^o i + sin10^o j)

The velocity of C relative to B is  v_C/B = v_C - v_B

                    v_C/B = (-1.157 i + -5.5974 j) - v_B(-cos10^o i + sin10^o j)

Set the x component of this vector equal to zero and solve for v_B.

Part b)

To find the velocity of the belt so that the relative velocity of C with respect to B is a minimum, you can do a graphical or an analytical solution. 

The graphical solution is easier.

Make a graphical sketch of the vector addition of this equation (it will be a triangle):

                  v_C/B = v_C + (-v_B)

You will see that for v_C/B to be minimum it must be perpendicular to v_B.

This allows you to solve the triangle for v_B.

Alternatively, you can do it analytically.

Using the same triangle and the law of cosines you get:

          v_C/B² = v_B² + v_C² - 2v_Bv_C cos(88.32^o)

You want to find the value of v_B that minimizes v_C/B.

So find the derivative of v_C/B with respect to v_B.

Then set this equal to zero and solve for v_B.