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Black_prince [1.1K]
4 years ago
8

1. Consider three objects: Object A is a hoop of mass m and radius r; Object B is a sphere

Physics
1 answer:
Zarrin [17]4 years ago
7 0

a) Object C (the disk) has the greatest rotational inertia (\frac{3}{2}mr^2)

b) Object B (the sphere) has the smallest rotational inertia (\frac{4}{5}mr^2)

Explanation:

The moments of inertia of the three objects are the following:

1) For a hoop of negligible thickness, it is

I=MR^2

where M is its mass and R its radius. For the hoop in this problem,

M = m

R = r

Therefore, its moment of inertia is

I=(m)(r)^2=mr^2

2) For a solid sphere, the moment of inertia is

I=\frac{2}{5}MR^2

where M is its mass and R its radius. For the sphere in this problem,

M = 2m

R = r

Therefore, its moment of inertia is

I=\frac{2}{5}(2m)(r)^2=\frac{4}{5}mr^2

3) For a disk of negligible thickness, the moment of inertia is

I=\frac{1}{2}MR^2

where M is its mass and R its radius. For the disk in this problem,

M = 3m

R = r

Therefore, its moment of inertia is

I=\frac{1}{2}(3m)(r)^2=\frac{3}{2}mr^2

So now we can answer the two questions:

a) Object C (the disk) has the greatest rotational inertia (\frac{3}{2}mr^2)

b) Object B (the sphere) has the smallest rotational inertia (\frac{4}{5}mr^2)

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

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Velocity can best be described by which of the following statements?
pogonyaev

Answer: Velocity can best be described as, the speed in a given direction.

Explanation: To find the answer, we need to know more about the Velocity of a body.

<h3>What is Velocity of a body?</h3>
  • Velocity is the rate of change of displacement.
  • It's a vector quantity and is measured in m/s.
  • It can be positive, negative or zero.
  • A body is said to be in uniform motion, then its velocity remains constant.
  • Change in velocity can be a change in speed.
  • The magnitude of velocity is less than or equal to speed.

Thus, we can conclude that, the option C is best describing velocity.

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2 years ago
Read 2 more answers
In a meeting of mimes, mime 1 goes through a displacement d1 = (6.00 m)i + (5.74 m)j and and mime 2 goes through a displacement
Blizzard [7]

a) Vector product: |d1 × d2| = (34.2 m) k

b) Scalar product: d1 · d2 = -0.874 m

c) (d1 + d2) · d2 = 16.1 m

d) Component of d1 along direction of d2: -0.21 m

Explanation:

a)

In this part, we want to calculate

|d1 × d2|

Which is the vector product between the two displacements d1 and d2.

The two vectors are:

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

The vector product of two vectors (a_x,a_y,a_z) and (b_x,b_y,b_z) is also a vector which has components:

r=(r_x,r_y,r_z)\\r_x=a_yb_z-a_zb_y\\r_y=a_zb_x-a_xb_z\\r_z=a_xb_y-a_yb_x

We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:

r_z=(6.00)(2.9)-(5.74)(-2.92))=34.2 m

Therefore, the vector product of d1 and d2 is:

|d1 × d2| = (34.2 m) k

b)

In this case, we want to calculate

d1 · d2

Which is the scalar product between the two displacements.

The scalar product of two vectors (a_x,a_y,a_z) and (b_x,b_y,b_z) is a scalar given by:

a \cdot b = a_x b_x + a_y b_y + a_z b_z

In this problem,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

Therefore, the scalar product between the two vectors is:

d_1 \cdot d_2 = (6.00)(-2.92)+(5.74)(2.9)=-0.87m

c)

In this  case, we want to calculate

(d1 + d2) · d2

Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.

Given two vectors  (a_x,a_y,a_z) and (b_x,b_y,b_z), the resultant vector is also a vector given by

r=(r_x,r_y,r_z)\\r_x=a_x+b_x\\r_y=a_y+b_y\\r_z=a_z+b_z

In this case,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

So the resultant vector is

r_x=6.00+(-2.92)=3.08 m\\r_y = 5.74+2.9=8.64 m

So

(d_1+d_2)=(3.08 m,8.64 m)

And calculating the scalar product with d2, we find:

(d_1 + d_2)\cdot d_2 = (3.08)(-2.92)+(8.64)(2.9)=16.1 m

d)

The component of a vector a along another vector b is given by

a_b = \frac{a\cdot b}{|b|}

wherea\cdot b is the scalar product between and b

|b| is the magnitude of vector b

In this problem, we have the two vectors

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

We want to find the component of d1 along the direction of d2.

We already calculated the scalar product of the two vectors in part b):

d1 · d2 = -0.874 m

The magnitude of a vector b is given by

|b|=\sqrt{b_x^2+b_y^2+b_z^2}

So, for vector d2,

|d_2|=\sqrt{(-2.92)^2+(2.9)^2}=4.1 m

Now we can calculate the component of d1 along d2:

d_1_{d_2}=\frac{d_1 \cdot d_2}{|d_2|}=\frac{-0.874}{4.1}=-0.21 m

Learn more about operations with vectors:

brainly.com/question/2927458

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brainly.com/question/1592430

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