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Genrish500 [490]
1 year ago
10

In a similar rolling race (no slipping), the two objects are 2 solid cylinders of the same mass but different radii. which reach

es the bottom first?
Physics
2 answers:
Brums [2.3K]1 year ago
7 0

The solid cylinder reaches the bottom first.

A solid object will always roll down the ramp faster than a hollow object of the same shape regardless of their exact mass or diameter.

The solid cylinder has a lower moment of inertia than the hollow one does. Although they have the same mass, all the hollow cylinder's mass is concentrated around its outer edge so its moment of inertia is higher.

The velocity would be greater for the material with radius of gyration k is lower of for which the moment of inertia is lower. Solid cylinder with k as 1/2 will have greater velocity as compared to hollow cylinder which has k as 1. As they have same mass and radius their energies at both the levels would remains same including the potential energy and energy lost due to friction.

learn more about moment of inertia from here: brainly.com/question/28167913

#SPJ4

Alika [10]1 year ago
5 0

The solid cylinder reaches the bottom first.

A (solid) cylinder is a solid that has two parallel planes and a cylindrical surface surrounding it. An element of the cylinder is a line segment that is determined by an element of the cylindrical surface between two parallel planes. A cylinder's components are all the same length. The material's velocity would be higher if the moment of inertia was lower and the radius of gyration k was smaller. In comparison to a hollow cylinder with k as 1, a solid cylinder with k as 1/2 will move at a faster speed. Their energies at both levels, including potential energy and energy lost through friction, would stay the same because they had the same mass and radius.

Learn more about solid cylinders here brainly.com/question/18801887

#4269

A solid cylinder and a hollow cylinder of the same mass and radius, both initially at rest, roll down the same inclined plane without slipping. Choose the correct options.

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A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

f=mg\ sin\theta

f=1100\ kg\times 9.8\ m/s^2\ sin(4)

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0
3 years ago
What is the genotype of the male?
shtirl [24]
The genotype of female is XY
3 0
3 years ago
A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.
Aloiza [94]

Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

tn = 20 seconds

A) In harmonic motion, we know that;

ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

T = 22/37.

Thus ;

ω = 2π/(22/37) = 10.5672

So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g

B) In simple harmonic motion, velocity is given as;

v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0
3 years ago
What is the rotational inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a
krek1111 [17]

We will apply the concepts of energy conservation to solve the problem. We know that gravitational potential energy is equivalent to the sum of translational kinetic energy and rotational kinetic energy. Additionally, the relation of the angular velocity with the tangential velocity will be determined to eliminate the angular term and obtain the expression of the Inertia in terms of the data given, therefore, we know that

PE_{g} = KE_{trans} +KE_{rot}

mgh = \frac{1}{2} mv^2+\frac{1}{2}I\omega^2

The angular velocity in terms of tangential velocity and radius is defined as,

\omega = \frac{v}{R}

Replacing,

mgh = \frac{1}{2} mv^2+\frac{1}{2} I(\frac{v}{R})^2

Multiplying by R^2,

gh(mR^2) = \frac{v^2}{2} (mR^2)+\frac{v^2}{2}I

\frac{v^2}{2}I = (gh-\frac{v^2}{2})mR^2

I = (\frac{2gh}{v^2}-1)mR^2

Replacing with our values we have,

I = (\frac{2(9.8)(2)}{6^2}-1)mR^2

I = 0.089mR^2

6 0
3 years ago
Which scientist used Tycho Brahe’s data to propose three laws of planetary motion?
slava [35]
That was his student, Johannes Kepler.
3 0
4 years ago
Read 2 more answers
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