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sukhopar [10]
3 years ago
10

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 14.8 rad. Du

ring that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity
Physics
2 answers:
AlekseyPX3 years ago
5 0

Answer:

(a) 1.184 rad/s²

(b)  2.96 rad/s

Explanation:

(a)

Using,

Ф = ω₀t + 1/2αt²....................... Equation 1

Where Ф = angular distance, ω₀ = initial angular velocity, α = angular acceleration, t = time.

Since the disk starts from rest, ω₀ = 0 rad/s,

Therefore,

Ф = 1/2αt²

make α the subject of the equation

α = 2Ф/t²............................ Equation 2

Given: Ф = 14.8 rad, t = 5.0 s

Substitute into equation 2

α = 2×14.8/5²

α = 29.6/25

α = 1.184 rad/s²

(b)

Using,

Ф = ω't...................... Equation 3

Where ω' = Average angular velocity.

Then,

make ω' the subject of the equation

ω' = Ф/t........................Equation 4

Given: Ф = 14.8 rad, t = 5.0 s

Substitute into equation 4

ω' = 14.8/5

ω' = 2.96 rad/s

11Alexandr11 [23.1K]3 years ago
3 0

Answer:

Explanation:

Given:

Initial θ = 0 rad (from rest)

Final θ = 14.3 rad

Time, t = 5 s

B.

Angular velocity, w = dθ / dt

= (14.3 - 0)/5

= 2.86 rad/s

A.

Acceleration, ao = dw/dt

Initial angular velocity, wi = 0 rad/s (from rest)

Final angular velocity, wf = 2.86 rad/s

a = (2.86 - 0)/5

= 0.572 rad/s^2

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A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
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Answer:

The frictional force acting on the block is 14.8 N.

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Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
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Answer:

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T = M_{a} *a    ..........(equation 2)

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a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

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