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14

You are right. To open the door during the same time interval (for pushing at $a$ and $b$), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives $$ \frac{F_a}{F_b} = \frac{r_b}{r_a} $$ where $r$ is the distance from the point of contact to the axis. However the ...


8

The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied But let's ...


7

Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is? When you push a body it will always rotate around the center of mass (white arrows) if you apply a force at the handle you are ...


5

To a first-order effect, there would be no change. But one consequence of melting is that the water moves to other places. Water that moves from the poles to other areas on the surface of the earth would serve to (slightly) increase the moment of inertia of the planet. This is because the mass of the water would be farther from the rotational axis. The ...


3

The answer is NO. Change of energy is work i.e $W = \Delta E$ and here the work done is $$W = \text{Torque} \cdot \text{angular displacement}$$ which is equal in both the cases. The only change is one needs to apply more force to achieve the same amount of torque at a smaller radius. $$\text{Force at "b"} > \text{Force at "a"}$$ but not the work done or ...


2

When you calculate $\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$ of a particle of mass, m, having a linear momentum of $\vec{\mathrm{p}}$ in an inertial frame via a rotating frame or rotating body where the acceleration is directed towards the origin, you get ...


2

There are two senses to $v_{\mathrm{min}}$. Firstly it means that the speed of the particle must be greater or equal to $v_{\mathrm{min}}$ at $\theta = \pi$ or else it will not complete the cycle. Instead it will follow a ballistic trajectory until such time that tension comes back into the string. Secondly, the speed of the particle is not constant: it is ...


2

First, two remarks: The melting of the ice in the Arctic ocean would have zero effect. What matters is the ice over Greenland and Antarctica. There is a very long-term secular slowing of the Earth's rotation rate due to the recession of the Moon from the Earth. This answer ignores that effect. (Alternatively, this answer has this secular effect subtracted ...


1

Torque $\tau = \vec{r} \times \vec{f} = I \vec{\alpha} $ $r$ here is moment of arm, ie. the distance(& perpendicular distance) from the axis of rotation of the body. You can only have and define torque if you have an axis about which the subject will rotate. It will be an absolute force driven motion, with NO torques compelled to induce.


1

As @Gautham said, W = Torque * angular displacement which is also equal to the energy needed. here the angular displacement is same as we know and torque will be also same(Large distance implies less force needed,less distance implies large force needed but torque will be same) So energy needed will be same in both cases.


1

The ball rolls because: the friction present between the surface pushes the atoms/particles in contact with surface or ground backwards, hence with centre of mass moving in a direction, the ball's bottom surface(ie. The surface in contact) moves in opposite direction or backwards, causing the ball to roll. The ball slips because there is no friction that ...


1

Your assumption that the ball will be faster if it rotates faster has a very logical explanation and is completely valid. If your ball has a circumference C it will cover that distance with each rotation. Let us say your C was 1m and the period was 2 rps (revolutions per second) then your ball would cover 2 metres in 1 second. If your period is doubled to be ...


1

Consider two different regimes. First - when the ball is still slipping, the relative motion of the ground and ball causes a force on the ball which (a) slows down the center of mass and (b) increases the angular speed. Once the ball rolls without slipping this force disappears. Second - when the ball does not slip, the contact point must be stationary. ...


1

The load increased but the input power driving the fan remained same. Moreover in very accurate measurements, the air drag can also not be neglected, all this will hold for a very another reason that the blades rotated by the motor are of very much comparable mass to the taped rubber & stuff.


1

By adding the rubber band you did two things: increased the air drag unbalanced the fan The bearings of a fan don't like to be unbalanced - the friction goes up significantly because as the fan picks up speed there will be a large lateral force (centripetal force keeping the rubber band plus object in their circular orbit). When you balance the fan it ...


1

If we treat the Earth as an isolated system then both its linear and angular momenta will remain constant. To answer your question you need only consider the angular momentum. The angular momentum is given by: $$ L = I\omega $$ Since $L$ is a constant, if the moment of inertia changes from $I_1$ to $I_2$ then we have: $$ I_1\omega_1 = I_2\omega_2 $$ and ...



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