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Bond [772]
3 years ago
14

Starting from a state of no rotation, a cylinder spins so that any point on its edge has a contant tangential acceleration of 3.

1 m/s2. You keep track of the fractional number of turns N1 made made after 2.9 s and then that N2 made after 10 s. (Those numbers do not have to be integers.) What should be the approximate ratio N2/N1 ?
Physics
1 answer:
Leni [432]3 years ago
3 0

Answer:

Explanation:

tangential acceleration at = 3.1 m  / s²

angular acceleration = tangential accn / radius

= 3.1 / r , r is radius of the cylinder .

IF N₁ be no of rotation in time t

θ = 1/2 α t² , α is angular acceleration , θ is angle in radian covered in time t

2π N₁ = 1/2 (3.1 / r ) x 2.9²

N₁ = 2.0757 / r

similarly we can calculate

2πN₂ = 1/2 (3.1 / r ) x 10²

N₂ = 24.68 / r

N₂ / N₁ = 11.89

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A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

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\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

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A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identi
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Explanation:

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