Q48E - University Physics with Modern Physics

Question

A flat (unbanked) curve on a highway has a radius of 170.0 m. A car rounds the curve at a speed of 25.0 m/s . (a) What is the minimum coefficient of static friction that will prevent sliding? (b) Suppose that the highway is icy and the coefficient of static friction between the tires and pavement is only one-third of what you found in part (a). What should be the maximum speed of the car so that it can round the curve safely?

Step-by-Step Solution

Answer

a) The coefficient of static friction is, $\mu =170.0m$ .

b) The maximum speed of the car is, $v=25.0m/s$ .

Step 1: Identification of given data

The given data can be listed below as,

 

• The radius of curve is, $r=170.0m$ .
• The speed of the car is, $v=25.0 m/s$ .

Step 2: Significance of centripetal force.

The centripetal force is a force that allows to move an object in curved path. The centripetal force always acting towards the center therefore, object can move easily in a curved path.  

Step 3: Determination of coefficient of static friction.

Part (a)

 

The centripetal force is given as-

 

 $F=\frac{m{v}^2}{r}$

 

Here   is the mass of object,  v is the speed of the car, r  is the radius of curve on highway.

 

Substitute 25.0m/s  for v , and 17.0 m  for r  in the above equation.

 

 $F=\frac{m{\left(25.0 m/s\right)}^2}{\left(170.0m\right)}$

 

The expression for the force of friction can be expressed as,

 

 $F=\mu mg$

 

Here $\mu$  is the coefficient of static friction,  g is acceleration due to gravity.

 

Substitute $\frac{m{\left(25.0 m/s\right)}^2}{\left(170.0 m\right)}$  for F ,  $9.8 m/s^2$ for  g in the above equation.

 

 $$\begin{gathered} \frac{m{\left(25.0 m/s\right)}^2}{\left(170.0m\right)}=\mu m\left(9.8 m/s^2\right) \\ \mu =\frac{{\left(25.0 m/s\right)}^2}{\left(170.0m\right)}\times \frac{1}{9.8 m/s^2} \\ \mu =0.36 \end{gathered}$$

 

Hence, required coefficient of static friction is, 0.36.

Step 4: Part (b)

The expression for the frictional force can be expressed as,

 

 $$\begin{gathered} \frac{m{v}^2}{r}=\mu m g \\ {v}^2=\mu g r \end{gathered}$$

 

Substitute $\frac{0.36}{3}$  for $\mu$ ,  $9.8 m/s^2$ for g , and 170.0m  for r  in the above equation.

 

 $$\begin{gathered} v^2=\frac{0.36}{3}\left(9.8m/s^2\right)\left(170.0m\right) \\ v=\sqrt{199.92} \\ =14.13 m/s \end{gathered}$$

 

Hence, the required speed is, 14.13 m/s .