A motorcyclist takes a turn with a radius of $500 m$. Both the mass of the engine and the motorcyclist is $160 kg$ and their centre of mass lies $0, 5 m$ above the ground when it is perfectly vertical. His tangential speed is $60 m / s$ and its tangential acceleration is $4 m / s^2$.

a) Make a clear drawing in top view and back view with the

speeds, accelerations and forces that affect the engine and motorcyclist.

b) Calculate the speed and acceleration of the motorcyclist.

c) Calculate the size of the components of all forces that act on

the contact surface of the engine with the road.

d) Calculate the angle at which the motorcyclist takes the turn (Hint: take its

mass centre as the origin of the system).

e) What is the apparent weight of the motorcyclist when he turns?*

What I have done and what I do not know:

a)

I was not sure about what **'top view and back view'** meant so I did the following:

This is the diagram analysed from above: https://imgur.com/a/JtpNmfr

This is the diagram analysed from the ground: https://imgur.com/a/9AwOrWa

Note that the forces have just been drawn on the latest diagram because of clarity purposes.

**Is there anything missing on the diagrams?**

b)

NOTE: I assumed uniform circular motion. **It makes sense for you what I did?**

**EDIT**

I was wrong assuming uniform circular motion because there is tangential acceleration; **the motorcycle does not move with uniform speed.**

c)

Here we just have to use Newton's second law:

$$\sum F = Ma$$

As we are under uniform circular motion:

$$F = M a_c = M \frac{u^2}{OB}$$

However, we are not asked for calculating the centripetal force but the ones exerted on the road and the engine (always including the person's mass). These are:

The force exerted on the road by the engine:

$$F = Mg = -1569,6 N$$

The force exerted on the engine by the road:

This is simply the reaction force (Newton's third law):

N = 1569,6 N

**This seems too simple, am I missing something?**

d)

I do not know how to calculate the angle properly. Actually I got:

$$\alpha = \frac{ut}{OB}$$

But depends on time. **Should I use kinematics to get the time and therefore the angle?**

e)

**Isn't the apparent weight just the normal force exerted on the motorcyclist? (i.e N = 1569,6 N)?**

'No I do not agree with your calculation of normal force $F_N$ and (radial) friction force $F_f$. You should have $F_N=W$ because these are the only vertical forces - there is no $cos \theta$ factor. Also $F_f=Ma_c$ because this is the only radial force - there is no $sin \theta$ factor.'

As the diagram states, $F$ is the sum of $N$ and $F_f$. I would say that for this to be true, we should have:

$$F = Fsin \theta + Fcos \theta$$

Having:

$$Fsin \theta = F_f = F_c$$

$$Fcos \theta = F_N = -W$$

Do you agree now?

By the way, What's $F$? On the video you shared, is treated as another reaction force (it is given the letter $R$). OK this is a silly question. R is just the tilted reaction, that is why the following equation holds:

$$Fcos \theta = F_N = -W$$

CALCULATIONS

$$F_N = Mg = 1569.6 N$$

$$f_r = \frac{M v^2}{r} = 1152 N$$

The tangential frictional force:

$$f_t = Ma_t = 640 N$$

Thus, the resultant force is the vector sum of the two frictional forces:

$$\sum F = 1317.84 N$$

Do you agree?

NOTE: I was wrong thinking that $\sum F=0$. However, the angular momentum is conserved due to $\sum \tau=0$

edited Dec 2 by JD_PM