Author Topic: Dekoyl's Questions (Read 21776 times) Tweet Share

dekoyl · « **Reply #30 on:** March 29, 2009, 11:03:25 pm »

Hmm the last one for tonight hopefully.

How is the answer to part b that?

Mao · « **Reply #31 on:** March 29, 2009, 11:20:24 pm »

net acceleration is -a

that means, $F_{up} - F_{down} = -aM$

if you want to deal with the entire system: then the driving force is the only uphill force, the downhill forces are friction (on both objects) and pull due to gravity (on both objects)
$\implies F_{traction} - (21000\sin 23^o + 17000\sin 23^o + 890 + 920) = -a \times (1700 + 2100)$

if you want to deal with the car only (gives the same answer), then the driving force is the only uphill force, the downhill forces are friction (on car only), pull by gravity (on car only), and tension
$\implies F_{traction} - (7500 + 17000\sin 23^o + 890) = -a \times (1700)$

to calculate net acceleration, find the forces on the trailer, going uphill is the tension force, going downhill is friction and gravity
$\implies -a = \frac{7500 - (21000\sin 23^o + 920)}{2100}$

dekoyl · « **Reply #32 on:** March 29, 2009, 11:37:53 pm »

Ah awesome. Thanks Mao =]

dekoyl · « **Reply #33 on:** March 30, 2009, 12:11:00 am »

Gah.. another one.

Part b again. Should it be $m\triangle v = 0.156(26 - 40)$ because it's force on bat BY ball? Solutions has 1.2(12.7-4.12) ie. the bat

/0 · « **Reply #34 on:** March 30, 2009, 12:27:19 am »

Force on bat by ball = Force on ball by bat, just in opposite directions (Newton 3).

Also, if you want the impulse on the ball, it should be $0.156(26-(-40)) = 10.296kgms^{-1}$ , since you have to take into account the direction of the initial and final velocity vectors.

dekoyl · « **Reply #35 on:** March 30, 2009, 12:50:14 am »

Quote from: /0 on March 30, 2009, 12:27:19 am

Force on bat by ball = Force on ball by bat, just in opposite directions (Newton 3).

Also, if you want the impulse on the ball, it should be $0.156(26-(-40)) = 10.296kgms^{-1}$ , since you have to take into account the direction of the initial and final velocity vectors.

Yeesh

Forgot all about that. Not good.

Anyway thanks saviour /0.

dekoyl · « **Reply #36 on:** April 16, 2009, 08:49:03 pm »

Let $d$ =greatest horizontal distance of ball. Find the maximum vertical height to which the ball can be thrown in terms of $d$ . Assume the initial speed is the same in either case.

Thanks

TrueTears · « **Reply #37 on:** April 16, 2009, 08:57:44 pm »

Assuming parabolic path.

$d = \frac{u^2sin(2\theta)}{g}$

solving for $u^2$ leads to $\frac{dg}{sin(2\theta)}$

sub this in $h_{max} = \frac{u^2sin^2(\theta)}{2g}$ yields

$h_{max} = \frac{\frac{dg}{sin(2\theta)}sin^2(\theta)}{2g}$

$sin(2\theta) = 2sin(\theta)cos(\theta)$

subbing this in : $h_{max} = \frac{d}{4}tan(\theta)$

dekoyl · « **Reply #38 on:** April 19, 2009, 06:51:22 pm »

^Thanks TT

How would one prove that the maximum range of a projectile is launched at $45^{o}$ ?

Flaming_Arrow · « **Reply #39 on:** April 19, 2009, 06:57:51 pm »

we know that range is $\frac{u^2 \sin (2\theta)}{g}$

subbing in $\theta = 45^o$

we get

$\frac{u^2 \sin (90)}{g}$

$\sin(90) = 1$ which is the largest sin value

then we get $\frac{u^2}{g}$

$\therefore$ $45^o$ is the angle that yields the maximum range

dekoyl · « **Reply #40 on:** April 19, 2009, 07:10:54 pm »

Quote from: Flaming_Arrow on April 19, 2009, 06:57:51 pm

we know that range is $\frac{u^2 \sin (2\theta)}{g}$

The book, if I remember correctly, does not use the formula $\frac{u^2 \sin (2\theta)}{g}$

They derive it and I figured how they did it

They did some implications as well, though.

Thanks FA

Mao · « **Reply #41 on:** April 19, 2009, 09:31:42 pm »

vertical velocity: $u_v = u\sin \theta$
horizontal velocity: $u_h = u\cos \theta$

time to take to get to the top: $t_{1/2} = \frac{u_v}{g} = \frac{u\sin \theta}{g}$

$R = 2\cdot t_{1/2}\cdot u_h = \frac{u^2 (2\sin \theta \cos \theta)}{g} = \frac{u^2 \sin 2\theta}{g}$

$\frac{dR}{d\theta} = \frac{2u^2}{g}\cos 2\theta$ , $\frac{d^2R}{d\theta^2} = -\frac{4u^2}{g}\sin 2\theta$

at maximum range, $\frac{dR}{d\theta} = 0, \; \frac{d^2R}{d\theta^2} < 0,\; 0\le \theta \le \frac{\pi}{2}$

$\implies \theta = \frac{\pi}{4} = 45^o$

Flaming_Arrow · « **Reply #42 on:** April 20, 2009, 01:54:10 pm »

Quote from: Mao on April 19, 2009, 09:31:42 pm

vertical velocity: $u_v = u\sin \theta$
horizontal velocity: $u_h = u\cos \theta$

time to take to get to the top: $t_{1/2} = \frac{u_v}{g} = \frac{u\sin \theta}{g}$

$R = 2\cdot t_{1/2}\cdot u_h = \frac{u^2 (2\sin \theta \cos \theta)}{g} = \frac{u^2 \sin 2\theta}{g}$

$\frac{dR}{d\theta} = \frac{2u^2}{g}\cos 2\theta$ , $\frac{d^2R}{d\theta^2} = -\frac{4u^2}{g}\sin 2\theta$

at maximum range, $\frac{dR}{d\theta} = 0, \; \frac{d^2R}{d\theta^2} < 0,\; 0\le \theta \le \frac{\pi}{2}$

$\implies \theta = \frac{\pi}{4} = 90^o$

i think u mean $\implies \theta = \frac{\pi}{4} = 45^o$

dekoyl · « **Reply #43 on:** April 20, 2009, 03:58:41 pm »

^Heh didn't catch that.

Q30 from book.
A road is to be banked so that any vehicle can take the bend at a speed of $30ms^{-1}$ without having to rely on sideways friction. The radius of the curvature is 12m. At what angle should it be banked?

I think the book made a mistake in the answers but I'm not that familiar with this so I don't want to make any attempts of correcting the book

Thanks

Flaming_Arrow · « **Reply #44 on:** April 20, 2009, 04:08:25 pm »

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