Question

3 a racetrack surface has the shape of an inverted cone on which cars race in horizontal circles. for a steady speed of 29 m/s, to what distance d should a driver take her car, if she wishes to stay on a circular path without friction? (9 marks)\\(f_n\sin 48^0 = f_g\\) 2 marks\\(f_n\cos 48^0 = f_c\\)\\(\frac{f_n\sin 48^0}{f_n\cos 48^0} = \frac{f_g}{f_c} = \frac{mg}{\frac{mv^2}{r}} = \frac{rg}{v^2}\\) 3 marks\\(\tan 48^0 = \frac{(r)(9.8)}{29^2}\\) 2 marks\\(r = 95m\\)\\(\sin 48^0 = \frac{r}{d} = \frac{95}{d}\\) 2 marks\\(d = 130m\\)\\(***\\)if the car reduces its speed to 20 m/s, find the coefficient of static friction required to keep it moving along the same radius.

Explanation:

Part 1: Distance \( d \) for speed \( 29 \, \text{m/s} \) (already solved in the problem, but let's rederive for clarity)

Step 1: Analyze Forces

The normal force \( F_N \) has vertical component \( F_N \sin 48^\circ \) balancing weight \( F_g = mg \), and horizontal component \( F_N \cos 48^\circ \) providing centripetal force \( F_c = \frac{mv^2}{r} \).

Step 2: Divide Force Equations

Divide \( F_N \sin 48^\circ = mg \) by \( F_N \cos 48^\circ = \frac{mv^2}{r} \):
\( \tan 48^\circ = \frac{rg}{v^2} \)

Step 3: Solve for Radius \( r \)

Rearrange: \( r = \frac{v^2 \tan 48^\circ}{g} \)
Substitute \( v = 29 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \), \( \tan 48^\circ \approx 1.1106 \):
\( r = \frac{29^2 \times 1.1106}{9.8} \approx \frac{841 \times 1.1106}{9.8} \approx \frac{934.01}{9.8} \approx 95 \, \text{m} \)

Step 4: Relate \( r \) and \( d \)

From geometry, \( \sin 48^\circ = \frac{r}{d} \), so \( d = \frac{r}{\sin 48^\circ} \).
\( \sin 48^\circ \approx 0.7431 \), so \( d = \frac{95}{0.7431} \approx 130 \, \text{m} \)

Part 2: Coefficient of Static Friction (\( \mu_s \)) for \( v = 20 \, \text{m/s} \)

Step 1: Identify Forces

Now, friction \( f_s = \mu_s F_N \) acts up the slope (since speed is lower, car tends to slide down).
Vertical equilibrium: \( F_N \cos 48^\circ + f_s \sin 48^\circ = mg \)
Horizontal centripetal force: \( F_N \sin 48^\circ - f_s \cos 48^\circ = \frac{mv^2}{r} \)

Step 2: Substitute \( f_s = \mu_s F_N \)

Vertical: \( F_N \cos 48^\circ + \mu_s F_N \sin 48^\circ = mg \)
\( F_N (\cos 48^\circ + \mu_s \sin 48^\circ) = mg \) ...(1)

Horizontal: \( F_N \sin 48^\circ - \mu_s F_N \cos 48^\circ = \frac{mv^2}{r} \)
\( F_N (\sin 48^\circ - \mu_s \cos 48^\circ) = \frac{mv^2}{r} \) ...(2)

Step 3: Divide Equations (2)/(1)

\( \frac{\sin 48^\circ - \mu_s \cos 48^\circ}{\cos 48^\circ + \mu_s \sin 48^\circ} = \frac{v^2}{rg} \)

Step 4: Substitute Known Values

\( r = 95 \, \text{m} \), \( v = 20 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \):
\( \frac{v^2}{rg} = \frac{20^2}{95 \times 9.8} = \frac{400}{931} \approx 0.4296 \)

Let \( \cos 48^\circ \approx 0.6691 \), \( \sin 48^\circ \approx 0.7431 \). Substitute into the ratio:

\( \frac{0.7431 - \mu_s \times 0.6691}{0.6691 + \mu_s \times 0.7431} = 0.4296 \)

Step 5: Solve for \( \mu_s \)

Cross-multiply:
\( 0.7431 - 0.6691 \mu_s = 0.4296 (0.6691 + 0.7431 \mu_s) \)
\( 0.7431 - 0.6691 \mu_s = 0.2885 + 0.3203 \mu_s \)
\( 0.7431 - 0.2885 = 0.3203 \mu_s + 0.6691 \mu_s \)
\( 0.4546 = 0.9894 \mu_s \)
\( \mu_s \approx \frac{0.4546}{0.9894} \approx 0.46 \)

Answer:

s:

• Distance \( d \) (for \( 29 \, \text{m/s} \)): \( \boldsymbol{130 \, \text{m}} \)
• Coefficient of static friction (for \( 20 \, \text{m/s} \)): \( \boldsymbol{\approx 0.46} \)