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The lean of a motorcyclist

1 vote

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:


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?




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


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


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?


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?


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

asked Nov 24, 2018 in Physics Problems by Jorge Daniel (656 points)
edited Nov 30, 2018 by Jorge Daniel
c) Regarding this comment:

'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:

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


$$Fsin \theta = F_f = F_c$$

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

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:

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


$$F_N = Mg = 1569.6 N$$

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

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 $\sum \tau=0$
e) $F_N = 1569.6 N$
The diagram which I provided states that $F$ is the (vector) sum of $F_N$ and $F_f$. It is not *another* reaction force. Neglecting the tangential friction force (ie assuming tangential acceleration is zero) $F$ is the vector resultant (ie total) reaction force on the  bike from the road.

Your question does not ask you to calculate $F$ so it is not worthwhile to use it in calculations. It asks you for the values of the separate *components* of the reaction/contact forces.

Writing Newton's 2nd law vertically, horizontally (ie radially) and tangentially we have $$N-W=0$$ $$f_r=ma_c$$ $$f_t=ma_t$$ That is all you need to do.
Just for the sake of curiosity: the resultant force would simply be: $\sum F= f_r + f_t$.
Yes, assuming that is vector addition.

1 Answer

1 vote
Best answer

The back/front view should look like this :

The vertical forces $N$ and $W$ cancel out. The resultant force is a radial friction force $f$ acting towards the centre of the turning circle. This is the centripetal force.

Another friction force acts tangentially, increasing the tangential velocity. This force is coming out of the page of the back/front view Free Body Diagram.

answered Nov 25, 2018 by sammy gerbil (28,746 points)
selected Nov 30, 2018 by Jorge Daniel
NOTE: There are more new comments above.

Regarding this comment:

'Another friction force acts tangentially, increasing the tangential velocity. This force is coming out of the page of the back/front view Free Body Diagram.'

The fact that the tangential friction force increases the tangential velocity goes against my intuition. Actually, I thought about it as pointing against the tangential velocity (a friction force used to be thought as pointing against the movement direction)...
Friction forces oppose the tendency toward relative motion. The point of contact on the rear wheel would move backwards over the ground if there were no friction (the wheel turns without moving forward), so the friction force on the rear wheel is forwards. Without friction the point of contact onthe front wheel would move forwards because the cycle is being accelerated and the front wheel is not rotating fast enough - ie the 'no slip ' condition is broken because of the linear acceleration of the cycle. So the friction force on the front wheel is backwards.

The resultant friction force is forwards. It is the friction force which causes tangential acceleration. Without it the cycle would not accelerate.