a) The coefficient of friction is 0.605
b) The coefficient of friction would decrease
Explanation:
a) The motorbike moves along the turn with a uniform circular motion, where the centripetal force is provided by the force of friction between the tires and the road.
Therefore, we can equate the two forces, and we find the following equation:
[tex]\mu mg = m \frac{v^2}{r}[/tex]
where
[tex]\mu[/tex] is the coefficient of friction
m is the mass
g is the acceleration of gravity
v is the speed of the motorbike
r is the radius of the curve
In this problem, we have:
r = 400 ft (radius)
[tex]g=32 ft/s^2[/tex] (acceleration of gravity)
[tex]v=60.0 mph \cdot \frac{5280 ft/mi}{3600 s/h}=88 ft/s[/tex] (velocity)
And solving for [tex]\mu[/tex], we find the coefficient of friction:
[tex]\mu = \frac{v^2}{gr}=\frac{88^2}{(32)(400)}=0.605[/tex]
b)
The coefficient of friction we calculated in part a) is the coefficient of friction when the road is dry. However, when it rains or it snows, the value of the coefficient changes. In particular, the force of friction when the road is wet (or ice) is lower (there is less "interaction" between the tires and the road), and this means that the coefficent of friction is also smaller.
The consequence of this can be seen from the same equation we used before:
[tex]\mu mg = m \frac{v^2}{r}[/tex]
Since the value of [tex]\mu[/tex] is lower, the force of friction is lower, and so the velocity of the motorbike, v, must also be lower. If the motorbike keeps the same speed given in part a), then the frictional force will not be able to provide the required centripetal force in order to keep the motorbike in circular motion, and therefore the motorbike will go out of the turn.
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