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erastova [34]
4 years ago
9

If an object is rolling without slipping, how does its linear speed compare to its rotational speed?

Physics
1 answer:
RoseWind [281]4 years ago
3 0

Answer:

v = rw

Explanation:

When an object is rolling continuously without slipping, then every angle it rotates through, is equal to a distance the perimeter has rotated.

If the object completes 10 revolutions and takes a particular time, let's say t to complete it. The angular distance would then be 20 π rad, while its angular velocity will be 20 π/t

The circumference will somehow translate to the distance it covers, which is 20πr, this means that the speed is 20πr/t

So, like the question asked, the linear speed compared to angular speed is

v : w

20πr/t : 20πt, which can be simplified to

r : 1

In essence, v = rw

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What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?
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Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

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Where,

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Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0            -------------(1)

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Where h= Sin 22 = h/d= h/4.1

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or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 mVf^{2}= 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 mVi^{2}=0.5×200×Vi^{2}=100Vi^{2} j

Put these values in eq 1, we get;

0-100 Vi^{2}+3018.4-0=0

-100 Vi^{2}=-3018.4

==> Vi^{2}= \frac{3018.4}{100} = 30.184

==>  Vi = \sqrt{30.184}  = 5.5 m.sec

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Graph B represents the velocity of the sphere changes over time when falling with constant acceleration.

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To know more about  constant acceleration. visit : brainly.com/question/9754169

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