Question

An SUV has a height \(h\) and a wheelbase of length \(b\). Its center of mass is midway between the wheels and at a distance \(\alpha h\) above the ground, where \(0<\alpha<1 .\) The SUV enters a turn at a dangerously high speed, \(v\). The radius of the turn is \(R(R \gg b),\) and the road is flat. The coefficient of static friction between the road and the properly inflated tires is \(\mu_{\mathrm{s}}\). After entering the turn, the SUV will either skid out of the turn or begin to tip. a) The SUV will skid out of the turn if the friction force reaches its maximum value, \(F \rightarrow \mu_{s} N\). Determine the speed, \(v_{\text {skid }}\), for which this will occur. Assume no tipping occurs. b) The torque keeping the SUV from tipping acts on the outside wheel. The highest value that the force responsible for this torque can have is equal to the entire normal force. Determine the speed, \(v_{\text {tip }}\), at which tipping will occur. Assume no skidding occurs. c) It is safer if the SUV skids out before it tips. This will occur as long as \(v_{\text {skid }}

Short answer

Short Answer: The skid speed (\(v_{skid}\)) of the SUV is given by the formula \(v_{skid} = \sqrt{\mu_s gR}\), while the tip speed (\(v_{tip}\)) is given by \(v_{tip} = \sqrt{\frac{b \mu_s g}{2 \alpha h}}\). For the SUV to skid before it tips, the maximum value of the center of mass' height (\(\alpha\)) should satisfy the condition \(\alpha_{max} = \frac{b}{2Rh}\).

Step-by-step solution

(Step 1): Analyzing the forces

To handle this situation, let's draw a free-body diagram of the SUV from the side view while it's turning. There are three forces acting on the SUV: gravity (\(mg\)), normal force (\(N\)), and static friction force (\(F\)). When the SUV reaches the verge of skidding, the static friction force is at its maximum value, which is \(F = \mu_s N\).

(Step 2): Applying Newton's second law

Now, let's apply Newton's second law in the horizontal direction: $$m\frac{v_{skid}^2}{R} = F = \mu_s N$$ In the vertical direction, we have: $$mg = N$$

(Step 3): Solving for v_skid

Combining the equations obtained in Step 2, we get: $$m\frac{v_{skid}^2}{R} = \mu_s mg$$ And solving for \(v_{skid}\), we find: $$v_{skid} = \sqrt{\mu_s gR}$$ #b) Finding the tip speed v_tip#

(Step 1): Analyzing the forces and torques

Now let's analyze the tipping situation. When the SUV is about to tip, the entire normal force acts on the outer wheel. We need to consider the torques caused by the gravitational force and the centripetal force acting on the center of mass. The torque due to the centripetal force tries to tip the SUV, while the torque due to gravity resists the tipping.

(Step 2): Applying Newton's second law and torque equations

Let's apply Newton's second law in the horizontal direction for the center of mass: $$m\frac{v_{tip}^2}{R} = F$$ In the vertical direction, we have: $$mg = N$$ Now let's calculate the torques. The torque due to gravity (\(\tau_g\)) is: $$\tau_g = \frac{1}{2} b mg$$ The torque due to centripetal force (\(\tau_c\)) is: $$\tau_c = F \alpha h = \frac{m}{\mu_s} g \alpha h$$

(Step 3): Equating torques and solving for v_tip

When the SUV is about to tip, the torques balance each other: $$\tau_g = \tau_c$$ $$\frac{1}{2} b mg = \frac{m}{\mu_s} g \alpha h$$ Now, let's substitute the centripetal force from Step 2 and solve for \(v_{tip}\): $$\frac{1}{2} b mg = \frac{m}{\mu_s} g \alpha h = m\frac{v_{tip}^2}{R}$$ $$v_{tip} = \sqrt{\frac{b \mu_s g}{2 \alpha h}}$$ #c) Determining the maximum value for α#

(Step 1): Comparing skid speed and tip speed

The condition for the SUV to skid before it tips is: $$v_{skid} < v_{tip}$$ Substituting the expressions for \(v_{skid}\) and \(v_{tip}\) we found earlier: $$\sqrt{\mu_s gR} < \sqrt{\frac{b \mu_s g}{2 \alpha h}}$$

(Step 2): Solving for the maximum α value

To find the maximum value of \(\alpha\), let's square both sides of the inequality and rearrange it: $$\mu_s gR < \frac{b \mu_s g}{2 \alpha h}$$ $$\alpha > \frac{b}{2Rh}$$ Thus, the maximum value for \(\alpha\) is: $$\alpha_{max} = \frac{b}{2Rh}$$

Key concepts

Newton's Second Law

Understanding Newton's Second Law of Motion is pivotal in physics problem-solving. In essence, it states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. The law is beautifully represented by the equation:

\[ F = ma \]
Where F stands for the net force, m for mass, and a for acceleration. This equation is a cornerstone for analyzing motion problems, whether in linear or circular dynamics, as in the case of an SUV turning a corner at high speed.

For the exercise we're focusing on, we consider only the horizontal direction since the SUV is at risk of skidding due to horizontal acceleration. When the SUV is at the brink of skidding, the static friction force equals the net force causing the horizontal acceleration. Therefore, we can set:

\[ m\frac{v_{\text{skid}}^2}{R} = \thinspace \mu_s N \]
This expression utilizes Newton's second law by equating the force of static friction to the inertia of the SUV moving in a curve.

Static Friction

When it comes to preventing an object from sliding on a surface, static friction is the hero without a cape. It's the resisting force that comes into play when an object is subjected to an external force without any actual movement occurring between the object and the surface. This force can be viewed as a threshold that needs to be overcome for motion to start.

The maximum force of static friction, before sliding begins, is expressed as:

\[ F_{\text{max}} = \mu_s N \]
Here, \( \mu_s \) represents the coefficient of static friction between the surfaces in contact, while \( N \) is the normal force - the perpendicular force exerted by a surface on an object. In our SUV scenario, we're interested in the moment the SUV starts to skid because that's the point where the static friction force is at its maximum. The vehicle's speed at which skidding occurs, \( v_{\text{skid}} \), is a direct result of this force reaching its upper limit.

Torque and Equilibrium

Torque is a measure of the rotational force's tendency to cause an object to rotate. The magnitude of torque depends on where and how the force is applied. It's given by

\[ \tau = r \times F \]
where \( r \) is the distance from the pivot point to the point where the force is applied, and \( F \) is the force. Torque plays a crucial role when dealing with situations involving equilibrium, where an object is in a state of balance and has no net force or net torque acting on it.

In the context of an SUV approaching a tipping point while turning, we must consider the equilibrium of torques. This is because the SUV can tip over when the torque caused by the centrifugal effect due to high speed overcomes the stabilizing torque generated by the SUV's weight. Equilibrium is reached just before tipping occurs, where the torque due to gravity is balanced by the torque due to the centripetal force.

Our exercise examines the relation between the torque that resists tipping (due to gravity) and the torque trying to tip the SUV over (due to the centrifugal force acting at the center of mass). Recognizing this balance can help us determine the speed at which the vehicle will tip, \( v_{\text{tip}} \), under the assumption that there's no skidding, giving us deeper insight into the rotational dynamics at play.