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anastassius [24]

3 years ago

13

A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance

h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil.(a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

1 answer:

yan [13]3 years ago

8 0

Answer:

a) h'= 5/7 h , b) h ’= h

Explanation:

Let's use energy conservation for this exercise

Starting point. Upper left side

Em₀ = mg h

Final point. Lower left side

Emf = K = ½ m v² + ½ I w²

Em₀ = emf

mgh = ½ m v² + ½ I w²

Angular and linear velocity are related

v = r w

w = v / r

The moment of inertia of the marble that we take as a solid sphere is

I = 2/5 m r²

We substitute

m g h = ½ m v² + ½ 2/5 m r² (v / r)²

g h = ½ v2 (1 + 2/5)

v = √(g h 10/7)

This is the speed at the bottom of the bowl

Now let's apply energy conservation to the right side

a) right side if rubbing

Em₀ = K

Emf = U = mg h’

½ m v² = mg h’

h’= ½ (g h 10/7) / g

h'= 5/7 h

b) right side with rubbing

Em₀ = K

Emf = K + U = -½ I w² + m g h

Emf = -½ 2/5 m r² v² / r² + m gh

Em₀ = emf

½ v² = -1/5 v² + g h’

h’= (1/2 +1/5) (gh 10/7) / g

h ’= h

c) When there is friction, an energy of rotation is accumulated that must be dissipated, by local it goes higher

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Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

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Substituting t = 0.200 s, we find

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And so now we can find the tangential speed at t = 0.200 s:

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2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

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$a=\frac{v^2}{r}$

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v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

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7 0

3 years ago

You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are pres

Dominik [7]

Answer:

<em>The required constant friction force for the last 20 m is 6,862.8 N</em>

Explanation:

<u>Energy Conversion</u>

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

$E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J$

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

$E_1=K_1+W_1$

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

$W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J$

This means that the kinetic energy that remains when the car reaches ground level is

$K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J$

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

$W_2=F_{r2}\cdot x_2=137,256$

Solving for Fr2

$\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N$

The required constant friction force for the last 20 m is 6,862.8 N

3 0

3 years ago

Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.

kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, $4.67\times 10^7$

(b) The number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{3.82\text{ days}}$

$k=1.81\times 10^{-1}\text{ days}^{-1}$

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.81\times 10^{-1}\text{ days}^{-1}$

t = time passed by the sample = 12 days

a = initially number of radon atoms = $4.1\times 10^8$

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

$12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}$

$a-x=4.67\times 10^7$

Thus, the number of radon atoms will remain after 12 days is, $4.67\times 10^7$

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = $(4.1\times 10^8)-(4.67\times 10^7)$

The number of radon nuclei have decayed = $3.6\times 10^8$

Thus, the number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

5 0

3 years ago