# Tag Info

1

It is there, it's just hidden by the change of coordinates. Written in Cartesian coordinates, the kinetic energy is $$T=\frac12m\dot{x}_1^2+\frac12m\dot{y}_1^2+\frac12m\dot{x}_2^2+\frac12\dot{y}_2^2+\frac12I\left(\dot\theta_1^2+\dot\theta_2^2\right)\tag{1}$$ where the last term is the rotational kinetic enregy. If you let \begin{align} ...

0

The lesser surface would be in contact with the road, the lesser will be the friction produced. In your case, when tyre is inflated, the less surface will be there of tyre which could make contact with road or track during motion. When our bike's tyre gets punctured it even don't move and this is due to the dilation of tyre surface which could make contact ...

0

Friction is more if the Area of Contact of the tire is more. When Air is completely filled in a Tire, the tire will be in lesser contact with the land as compared to when it has no or less air in it. This will lead to more friction by the tire, which will make it more difficult to drive the bicycle as compared to the earlier tire.

1

You can get answer to your first question here. When tires are inflated less area of the tire comes in contact with the ground, and thus there is less kinetic friction.

1

The only outright requirement is that you compute all the angular momenta in your problem around the same center (modulo applying the parallel axis theorem to break the angular momentum of extended bodies into of-and-around the CoM parts). So you can freely chose any single point to use Now, as with most such "free" choices in physics there are generally ...

1

It is simply not true, as stated. A cube is as perfectly balanced around its center of mass as a sphere is. You have shown it mathematically, and you are perfectly correct. However, the principal axes may be chosen perpendicular to the faces of the cube. Of course, when the point is the center of mass, this is no better than any other choice, except for ...

8

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ...

0

Both graphs have 0% force at 0% slip. That implies that no force is being applied to the tire. As applied force is increased until a certain point the increase in slip remains nearly linear because the coefficient of friction remains nearly constant. Thus effective force also increases nearly linearly. After that point the coefficient of friction gets ...

2

All your thinking is very good and correct. BUT you are missing a point in the question :) Yes, the door's weight (which pulls from the center og gravity - no need to think about each particle of the door) creates a torque. The hinges then gives a counter torque. True. BUT let's read the question: A door is hinged at one end and is free to rotate about ...

0

The torque equation is $$\tau = (r \times F) = rF \sin \theta$$ $r$ is a vector from the point from which torque is measured to the point where force is applied. $\times$ is the cross product $\theta$ is the angle between the position vector and force. F is the amount of force directed perpendicularly to the position of the particle. Any force directed ...

2

The discussion is mostly semantic. They are both calculated relative to a point, in the case of the torque the point has the additional meaning that if you put an axle trough the point, the object will start to rotatte around it if the net torque is not zero. It happens also that the torque will be the same if you chose any other point along the axis ...

0

There can be no torque produced due to weight as they both are perpendicular to each other and therefore the work done by Gravitational force will be $zero$.

1

Since the direction of friction is specified by the direction of motion, it must be the opposite direction, right? Wrong. You can go back to whoever told you that and yell at them. :-P Seriously though: the direction of friction actually has nothing to do with the direction of motion. One really obvious way to see this is that you can make the ...

1

Using my understanding of definite integrals, $I_i$ would be the inertia of the mass that you are not including in your calculation, while $I_f$ is the inertia of all the masses, including your mass of interest. $I_f$ - $I_i$ therefore is the inertia of the mass you are interested in. It appears as if $I_f(m)$ and $I_i(m)$ are not a functions of mass per se ...

0

So a moving beam can be analyzed the same as a static beam, if the inertial forces are included. Consider the example you gave: Kinematics Track the center of mass starting from the position coordinates, and use chain rule of differentiation to get to velocity and acceleration. \begin{aligned} x &= \frac{\ell}{2} \cos \theta & y & = ... 0 If your rod is a pendulum that was initially held horizontal by a force at the free end, before release it was a beam like you know how to calculate. It wouldn't be quite straight, it would sag in the middle, creating stress and strain. When you release it, each small piece of the rod is acted on by gravity and the stresses in the rod, and accelerates ... 1 The idea is that if there are no forces on an object, then no matter how it rotates, its center of mass must move at a constant velocity. Then in the frame of the object, the center of mass appears stationary and everything else rotates around it. In general this cannot be said of any other point in the object. To see that the center of mass moves at a ... 2 It happens I think because rotation about axes which do not pass through the center of mass are usually inherently unstable and hence the rotation tends to "decay" to a more stable axis, i.e the one passing through its center of mass. Also, the moment of inertia of a body is usually the lowest through a certain principle axis through its center of ... 2 Expanding the correct answer of @SAKhan: Assume that the conical pendulum is rotating at an angle \theta at an angular velocity \omega. Note also that the radius of the circle is given byr=L\sin(\theta)$$For the point mass to move in a horizontal circle, the total vertical force is zero:$$T\cos(\theta)=mg$$The net horizontal force must supply ... 4 If you solve the problem for the two forces the vertical and the horizontal force (which is required for the circular motion) you obtain the relation$$\omega^2=\frac{g}{L\cos\theta}$$Hence the minimum required \omega is \sqrt{g/l}, the same as the angular frequency for motion of a bob in a plane. 0 There is such minimun agngular speed: You will always find an an angle that results in circular motion for any given angular speed. The angle is given by:$$\cos \theta=\frac{g}{L \omega^2}$$Update: I got to this expression by using the equations of motion: m\omega^2L \sin \theta=T\sin\theta and T\cos\theta=mg What this means is that the minimum ... 2 In the cases of Bottle A and Bottle C, they are full and empty. So the amount of water in them (or not) can be considered to be a complete system along with the bottle, since there is no possible way in which the fluid in the full bottle would reduce its volume or overall distribution, certain properties like the system's center of mass,center of gravity and ... 0 So we have than with a mixture of water and air we have more energy dissipation? One possible reason is this: the B bottle is assymetric, with the heavy watered side down. While it turns, water shpuld turn sides. Because of the viscosity of the water nd n relation with the glass, water has to exert a work, thus losing energy 0 You're talking about the longitudinal axis or "roll axis". Its value is the "bank angle" or "roll angle". Of course there's no limit on that. Its first derivative is the "roll rate". In fighter aircraft this is as high as 720 degrees/second. I can't think of any upper limit on this, because you could make the whole wing an aileron, and if you tilted it far ... 1 what does the commentator mean by concavity of the floor? He or she means that the surface you walk on is in fact the inside of a cylindrical surface. Like a very large version of the inside of a wedding-ring. The curvature of this surface can be measured. 1 That the floor is not flat but circular, an the centrifugal force acts radially, and is the same at the same radius. If you cover the concavity of the floor with a planar surface or either make the stations of planar segments (a polygon instead of a circle), then you will measure different accelerations at different points of the floor, because they are not ... 4 It is fully explained in this answer Angular momentum  L = m*v*r, since angular velocity is linear velocity divided by the radius, ω = v/r \rightarrow v = \omega *r , then L = \omega*[r*r*m] since L = \omega*I \rightarrow I = m *r^2 In the second sketch m=2, v = 3, r = 2 \rightarrow L = 2*3*2 = 12 Kg*m^2/s Since angular velocity is v/r = ... 1 The moment of inertia is linked with the kinetic energy.$$E_c=\sum_i \tfrac12 m_i v_i^2 = \sum_i \tfrac12 m_i (\omega r_i)^2 = \tfrac12 \omega^2 \sum_i m_i r_i^2$$(wikipedia) It seems quite logical to me. Other reason is dimensional. 1 First, there is no reason why a physical quantity must be linear in r. That only happens for some variables (perhaps the ones you are familar with). In the case of the moment of inertia, it is actually a multiplication of two linear variables in r, that is why it end result is quadratic in r. By definition: the moment of inertia is I=L/\omega, so ... 2 Nice question. First, look at the following Free Body Diagram. For the disk$$\text{T}\text{R}=\text{I}_{\text{cm}}\alpha $$where \text{T} is the tension in the string, \alpha is the angular acceleration in the disk and \text{I}_{\text{cm}} is the moment of inertia of the disk about it's center, which equals ... -1 If we look at the symmetry of the pear of mass m, we can see on the x axis that momentum is conserved because the derivative of its motion in space x is the same from any point we measure it's motion from:$$p_x=m \frac{dx}{dt}$$On the y axis however, the system is affected by an external field gravity, g which pulls the pear downwards. This removes ... 2 with your definition of dA you must integrate between r and 0, because you start at the center, and the rings grow in radius as you go for (r-x) from x=r to x=0 6 There is a general and simple formula to calculate the moment of inertia with respect to some axis if the moment of inertia with respect to another axis is known. I am pretty sure you'll be able to find it in your textbook. The theorem is parallel axis theorem. 5 What would happen if axis of rotation pass through centre of mass of an object? Will the object rotate when we will apply force to the object? For the sake of future readers, I'll reply to the original question: Every body has a Centre of Mass, whatever its form. If an object has a regular shape and uniform, homogeneous distribution of mass its ... 0 I assume that you are essentially asking whether an unhinged rod placed on a smooth surface will start rotating if a force perpendicular to it is applied at one of its ends. Yes it will. But why ? Why doesn't the rod as a whole start moving in a purely translational manner in a particular direction when you apply torque to one end ? You can think of it ... 1 Other sources do not point out that term I have problems with. Other sources explicitly assume a constant angular velocity and thus ignore that component. The wikipedia article you cited is correct. In any case, I want to know how you evaluate that derivative. Given any vector quantity \mathbf q that is the same (other than component ... 1 If you just follow your nose, then...$$\left(\frac{d}{dt}\right)_{rotating} {\bf{\Omega}} =\frac{d\bf{\Omega}}{dt}+\bf{\Omega}\times {\bf{\Omega}}$$Do you know what the second term is equal to? Hopefully this clears up the problem you have. -1$$ \mathbf V = \mathbf v + \boldsymbol \Omega \times \mathbf r $$The derivative of \mathbf V in the inertial frame is indeed,$$ \mathbf A = \frac{\mathrm D \mathbf v}{\mathrm Dt} + \frac{\mathrm D \boldsymbol \Omega}{\mathrm D t} \times \mathbf r + \boldsymbol \Omega \times \frac{\mathrm D \mathbf r}{\mathrm Dt}.$$You are right, both \mathbf v and ... 3 Definition of angular momentum A body B with velocity (and linear momentum) has a potential rotational momentum L with reference to/around any point/body O which does not lie on its trajectory. The magnitude of L can be found multiplying its linear momentum (p = m*v) by the distance of point O from the trajectory: r. In the full formula: L = m * [v * ... 3 I see now your problem and I believe that I can help. Let's begin from the velocity formula$$v_i = v_r + Ω \times r .\tag{1}$$Let's take the derivative of v_i IN THE INERTIAL frame,$$a_i = \left(\frac{dv_r}{dt}\right)_i + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times v_i . Here we use as much as we can our formula \$(dF/dt)_i = (dF/dt)_r + Ω ...

0

Can it be said that the net force pointing in the direction towards the center of the circle is equal to the centripetal force; or, as I seem to have mistakenly assumed, the net force on the object is equal to the centripetal force? Read the above sentences twice. I'll explain with respect to them. First, let us get the concept of centripetal force clear. ...

3

On a completely frictionless floor, with the absence of other external forces, the centre of mass of the car will continue in the same trajectory for ever. Hence no steering is possible. However, irrespective of whether the front wheels are rotating or not, turning of the front wheels will produce a counter torque changing the orientation of the car, albeit ...

27

Yes you can It is actually possible with a real car, but you would have to be very patient to steer a little bit. Suppose you have built a car with power on the big front wheels to induce a gyroscopic effect. If you rotate the wheels, the direction in which the center of mass is going will not change directly, but the angle in which the rest of the body ...

10

Since there is no friction, then it will not affect any other forces that may act on the car. The direction of wind blowing on the car may change its trajectory, as any driver will attest when driving in high winds. Turning the car wheels may have a slight affect on the resultant direction of the force. If the car has curved roof, then it may acts as ...

4

Friction is the only force that would cause the car to move along a different path. On a frictionless surface, the gyroscopic effect could change the orientation of the car a bit, but not the trajectory of the car. In other words, the front car would no longer point along the direction of travel, but would "skid". (That is, if you could call frictionless ...

47

If the wheels had spun fast enough for a gyroscopic effect to become noticeable, the only result on a frictionless surface (which would be the same without a surface at all) is that when you turn the wheels, the rest of the car would rotate instead of just the front wheels :) You need some reaction force to alter the trajectory, like a sail or surface ...

53

No, a car cannot steer on a frictionless surface. This has little to do with gyroscopic action and more to do with conservation of momentum: to turn, even when conserving its speed, the car needs to accelerate at right angles to its motion, which changes the total momentum of the motion. This change in momentum requires a force which, in normal roads, is ...

1

Your answer seems to be good. I solved it in the same way. However there are two points in which i have no clarity. The first thing i do not have clear is the conservation of energy in the problem. We use the energy conservation, but how do we guarantee that the normal force is conservative? And, still, in the conservation equations we do not consider its ...

0

If the person starts with an initial velocity equivalent to the tangent velocity at the edge of the station, he will not slide across the surface on the inside regardless of friction (friction is meaningless in this case, as it cannot act if the person has 0 velocity in respect to the floor beneath him). Imagine we are looking at this space station as a ...

0

Friction between the person's shoes and the floor/outer wall of the station would only exist if the person were attempting to start walking around the circumference of the station and thus participate in non-uniform circular motion. Think about how his bowling ball would behave if he just put it down at his side.

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