# Tag Info

1

Because your arm is effectively rotating, along with the ball, the latter will continue to rotate at the same speed after it leaves the hand. In reality, the fingers usually impart a spin though friction as the ball is released and slides out of the hand.

3

Moments of inertia are additive, so if you have lots of separate elements you can just sum up all the individual moments of inertia to get the total. In this case you can regard your door as being made up of lots of rods: OK it's a slightly odd looking door, but the point is that if the door is (conceptually at least) made up by stacking $N$ rods then ...

2

When calculating the moment of inertia of a mass along an (ideal) axis, the only thing that matters is how much mass is at any given distance from the axis. The extent and distribution of mass either parallel to the axis or circumferential to the axis is immaterial. Your example of a door (a rectangular prism) versus a rod is good; assuming the masses and ...

0

Yes, the distance between the axis of rotation and the center of mass is taken into consideration when talking about r. The tangential acceleration is always perpendicular to r. If however a force is applied at an angle to the system, the angle will have to be taken into consideration and we will consider the part of the force perpendicular to r. The angular ...

-1

Consider the diagram below: Shown are crank, chain wheel, chain, sprocket and rear wheel (drive wheel). Assume a constant force $F$ is applied to the end of the crank the question then is how does this result in an acceleration $a$ and what's the magnitude of $a$? The force $F$ causes a torque $T$ about the centre of the crank: $$T=FR_c$$ Now we can ...

-1

The first question you ask! Just think about the mass of the gear and how much torque is requiered to rotate it if there was no chain. With the chain, you need to apply enough torque so that the bike + driver system start moving and keep the movement in the case of friction too. You need force to accelerate objects, not to maintain their movement and ...

0

Following the link you gave in your question, torque steering as described there really happens in situations where the drive torque is applied to two wheels (front or rear), and for some reason the torque is not applied equally to both wheels. An apparent pull to the side when you accelerate your bike hard may be related to the tension in the chain, and ...

1

when people refer to a "powerful" car ... they actually mean acceleration. This means Torque (which gets translated to Force at the end of the drivetrain). And Force = m x a ... so for a given mass, Torque == Force == acceleration. Unfortunately, the technical definition of the term Power is defined as Torque x Revs. This means the Power curve is sort of ...

0

In your equation (1), $T = 8.02 m$ and in your equation (4) $-T\sin{21^{\circ}} + \mu mg - \mu T \cos{21^{\circ}} = 0$. Rearranging (4) and then substituting the value of T from (1) gives $$\mu = \frac{T\sin{21^\circ}}{mg - T\cos{21^\circ}}= \frac{8.02(0.358)}{9.8-8.02(0.934)}=1.24$$ There is a minus sign in the denominator. You must have added instead of ...

2

Let you apply force $\bf F$ at point $P$ the coordinate of which is $\bf r$ measured from a specific point $O$ - the point about which you want to rotate. Let $\bf r$ and $\bf F$ be in the same plane. Now, if you were to rotate $P$ about $O$, it would rotate around some axis perpendicular to the plane in which the force and the point lies; if ...

0

HINT: Torque on the rod about point on the ground, Since the rod is not rotating the $\tau_{net}=0$ $$Tl\cos69-mg\frac l2\cos42=0\\$$ Forces acting on the rod, Since the rod is not slipping $F_{net}=0$ $$N=mg-T\sin69\\ T\cos69-F_{friction}=0\\ F_{friction}=\mu N\\$$ $$F_{friction}=\mu(mg-T\sin69)\\ T\cos69-\mu(mg-T\sin69)=0$$ There you go now I ...

0

$\vec{v}=\vec{w}\times \vec{r}$ $\nabla \times \vec{v}=\nabla \times (\vec{w}\times \vec{r})$ $=\vec{w}(\nabla .\vec{r})-(\vec{w}.\nabla)\vec{r}$ $\nabla .\vec{r} = 3$ and $(\vec{w}.\nabla)\vec{r}=w.(\nabla\vec{r})=w$ therefore $\nabla \times \vec{v}=2w$ or see: \nabla \times \left( {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} ...

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