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Statement-1: \(\Lambda\) cyclist always bends inwards while negotiating a curve. Statement-2: By bending he lowers his centre of gravity.

Short Answer

Expert verified
Yes, Statement-2 explains Statement-1 as it provides the reason for the cyclist's inward lean.

Step by step solution

01

Analyze Statement-1

Statement-1 says a cyclist bends inwards while turning around a curve. This suggests he leans towards the center of the circular path he is following. This is typically done to balance the forces acting on him due to his circular motion and maintain equilibrium. The inward lean helps counteract the centrifugal force and prevents the cyclist from toppling over.
02

Analyze Statement-2

Statement-2 claims that the cyclist bends to lower his center of gravity. Lowering the center of gravity increases stability, as it allows for a lower pivot point and reduces the risk of toppling. This logically complements his need to maintain balance during the curve.
03

Understand the Relationship

Now, we need to analyze if Statement-2 provides a reason for Statement-1. When a cyclist bends inward while negotiating a curve, he does lower his center of gravity. This action provides greater stability and helps maintain balance, suggesting that Statement-2 explains why Statement-1 occurs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Center of Gravity
The center of gravity is a crucial point in any object or system, representing the average location of its weight distribution. In simpler terms, it is that spot where you could balance the object perfectly if it were on the tip of a pin. For cyclists, the center of gravity is generally in the lower torso. Lowering this point is essential in enhancing balance, especially when turning corners. When a cyclist negotiates a curve and bends inward, they're effectively lowering their center of gravity. This action allows them to have a better control of the bike. A lower center of gravity leads to a more stable position, which is vital when the forces acting on the bike change quickly, such as when navigating a curve. The closer the center of gravity is to the ground, the smaller the tendency of the bicycle to topple, hence providing the cyclist with a smoother and safer ride.
Centrifugal Force
Centrifugal force is not an actual force, but rather an apparent one experienced when an object moves in a circular path. As they move along a curve, cyclists need to contend with this force, which appears to push them outward, away from the center of the circular path. This is due to inertia wanting to keep the cyclist moving in a straight line and not naturally in a circle. To counteract this inertial effect, cyclists must lean inward, balancing the imaginary centrifugal force. This lean is achieved by actively shifting their center of gravity inwards, thus working against the outward momentum. This balance between the forces plays a vital role in ensuring that the cyclist can successfully and safely navigate curves without skidding or falling.
Equilibrium
Maintaining equilibrium is all about staying in a state of balance where opposing forces are equal. For a cyclist, equilibrium during a turn is achieved by the right combination of leaning and steering. But why exactly is this so important? In the context of circular motion, achieving equilibrium effectively means that the cyclist is neither falling inwards nor outwards. By leaning into the turn, cyclists can perfectly balance the forces of gravity and centrifugal force. This careful positioning allows the forces to cancel each other, enabling the cyclist to safely traverse the curve without tipping over. Staying in equilibrium not only enhances safety but also efficiency in energy use, permitting cyclists to maintain speed and control.
Stability
Stability is crucial for any moving object, especially on a bicycle that relies heavily on balance and control. A stable ride is achieved by maintaining a good posture and ensuring that the center of gravity is low and well-positioned. When turning through a curve, the cyclist's ability to maintain stability is significantly tested. Lowering the center of gravity, as mentioned earlier, enhances stability. Logic follows from basic physics principles: a lower pivot point provides a much stronger resistance against forces trying to topple the cyclist. With careful body positioning, the cyclist can enhance stability which allows for smoother navigation of turns, reducing the risk of accidents. Pedal positioning, grip on the handlebars, and centered weight distribution all contribute to maintaining stability, especially in the dynamic and sometimes unpredictable nature of cycling.

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Most popular questions from this chapter

A force \(F--\frac{k}{x^{2}}(x \neq 0)\) acts on a particle in \(x\) -direction. Find the work done by this force in displacing the particle from \(x=a\) to \(x=+2 a .1\) lere \(k\) is a positive constant.

A small block of mass \(m\) is kept on a rough inclined surface of inclination \(\theta\) fixed in an clevator. The elevator goes up with a uniform velocity \(v\) and the block does not slide on the wedge. The work done by the force of friction on the block in a time \(t\) will be: (a) zcro (b) \(m g v t \cos ^{2} \theta\) (c) \(m g v t \sin ^{2} \theta\) (d) \(\frac{1}{2} m g v \sin ^{2} \theta\)

A body of mass \(M\) was slowly hauled up the rough hill by a force \(F\) which at cach point was directed along a tangent to the hill. Work done by the force (a) is independent of shape of trajectory. (b) depends upon vertical component of displacement but independent of horizontal component. (c) depends upon both the components. (d) does not depend upon coefficient of friction.

Which of the following force is/are not conservative? (a) \(\vec{F}-a \hat{i}+b \hat{j}\) (b) \(\vec{F}-a x \hat{i}+b y \hat{j}(a \neq b)\) (c) \(\vec{F}-x y \hat{i}\) (d) \(\vec{F}-x^{2} y \hat{i}-2 x y^{2} \hat{j}\) (e) \(\vec{F}-\left(a x+b x^{3}+c x^{4}\right) \hat{i}\) (f) \(\vec{F}-A x^{2} \hat{i}+B x y \hat{j}\)

\(\Lambda\) chain of mass \(m\) and length \(l\) is held vertical, such that its lower end just touches the floor. I released from rest. Find the force exeried by the chain on the table when upper end is about to hit the foor. Solution Force \(F\) exerted by chain consists of two components (a) \(F_{1}\) weight of the fallen portion of the chain, (b) \(F_{2}\) thrust of the falling part of chain. Now consider an clement of chain of length \(d y\) at a height \(y\) from the floor. It will strike the floor with a velocity \(v-\sqrt{2 g y}\). Thus we have, \(\Gamma_{1}=\lambda y g\) Here \(\lambda\) is the mass per unit length of chain and \(\Gamma_{2}-v_{\mathrm{rel}} \frac{d m}{d t}\) We have \(v_{\mathrm{rel}}=v \quad\) and \(\quad d m=\lambda d x \quad \therefore F=-v \frac{\lambda d x}{d t}-\lambda v^{2}\) 'Ihe force exerted by chain on the floor,$$ F=F_{1}+F_{2}=\lambda y g+\lambda v^{2}-\lambda y g+\lambda(\sqrt{2 g y})^{2}=\lambda y g+2 \lambda g=3 \lambda y g $$ When upper end is about to hit the floor, \(y=l\) \(\therefore \quad F=3 \lambda / g=3 m g\)

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