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Leona [35]
3 years ago
11

A thin uniform cylindrical turntable of radius 2.7 m and mass 22 kg rotates in a horizontal plane with an initial angular speed

of 12 rad/s. The turntable bearing is frictionless. A clump of clay of mass 11 kg is dropped onto the turntable and sticks at a point 1.7 m from the point of rotation. Find the angular speed of the clay and turntable. Answer in units of rad/s. 010 (part 2 of 2) 10.0
Physics
1 answer:
Marizza181 [45]3 years ago
5 0

Answer:

8.59 rad/s

Explanation:

Given that:

A thin uniform cylindrical turntable radius = 2.7 m

with a mass (M) = 22 kg

initial angular speed (ω₁) =  12 rad/s

Mass of the clump of the clay (m) = 11 kg

Diameter (d) from the point of rotation = 1.7 m

We are to find the final angular  velocity  (ω₂) ,To do that; we apply the conservation of annular momentum; which is as follows:

L₁ = L₂

l₁ω₁ = l₂ω₂

( 0.5 × M × r²) × ω₁ = (0.5 × M × r² + md²) ω₂

Making ω₂ the subject of the formula ; we have:

\omega_2 = \frac{(0.5*M*r^2)* \omega_1}{(0.5*M*r^2+md^2)}

\omega_2 = \frac{(0.5*22*2.7^2)* 12}{(0.5*22*2.7^2+11*1.7^2)}

\omega_2 = 8.58 rad/s

Hence,  the angular speed of the clay and turntable = 8.59 rad/s

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A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the
emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

<u>d = 0.87 m</u>

5 0
3 years ago
We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with
amm1812

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m

6 0
3 years ago
A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

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6 0
2 years ago
A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o
Nitella [24]

Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

7 0
3 years ago
A transformer works when an alternating current flows through the primary coil. This produces an alternating magnetic field in t
neonofarm [45]

a voltage in the secondary coil

4 0
3 years ago
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