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kondaur [170]
3 years ago
14

You are sitting on a merry-go-round at a distance of 2m from its center. It spins 15 times in 3 min. What distance do you move a

s you make one revolution? What is your angular speed in RPM? What is your linear speed in meters per SECOND?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer:

A) 12.57 m

B) 5 RPM

C) 3.142 m/s

Explanation:

A) Distance covered in 1 Revolution:

The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:

s = rθ

where,

s = distance covered = ?

r = radius of circle = 2 m

θ = Angle = 2π radians  (For 1 complete Revolution)

Therefore,

s = (2 m)(2π radians)

<u>s = 12.57 m</u>

B) Angular Speed:

The formula for angular speed is given as:

ω = θ/t

where,

ω = angular speed = ?

θ = angular distance covered = 15 revolutions

t = time taken = 3 min

Therefore,

ω = 15 rev/3 min

<u>ω = 5 RPM</u>

C) Linear Speed:

The formula that gives the the linear speed of an object moving in a circular path is given as:

v = rω

where,

v = linear speed = ?

r = radius = 2 m

ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s

Therefore,

v = (2 m)(1.571 rad/s)

<u>v = 3.142 m/s</u>

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A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

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4 years ago
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