1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lemur [1.5K]
2 years ago
6

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its centre and perpendicula

r to its length. A mass m1 = 5.40 kg is attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
Physics
1 answer:
astraxan [27]2 years ago
7 0

Complete Question:

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis  attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?

(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

Answer:

a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J

Explanation:

a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.

As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.

The total rotational inertia will be as follows:

I = M*L²/12 + m₁*r₁² + m₂*r₂²

⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

⇒ I =  8.53 kg*m²

b)  The rotational kinetic energy of the rigid body composed by the rod and  the point masses m₁ and m₂, can be expressed as follows:

Krot = 1/2*I*ω²

if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:

Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²

⇒ Krot = 31.1 J

c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:

I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

I = 7.90 kg*m²

d) The new rotational kinetic energy will be as follows:

Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²

Krot= 28.8 J

You might be interested in
What is one characteristic of informal writing?
7nadin3 [17]
D. It is personal in style.
8 0
2 years ago
Read 2 more answers
Where is the fulcrum of a broom?
algol [13]

It's one of your hands.  Which one it is depends on how you sweep.

-- If you hold the top of the stick motionless and wave your bottom hand
back and forth, then your top hand is the fulcrum, and you're using the
broom as a Class-3 lever.

-- If you hold your bottom hand motionless and wiggle the top end of the
broom back and forth with your top hand, then your lower hand is the fulcrum,
and you're using the broom as a Class-1 lever.


8 0
3 years ago
Read 2 more answers
If the net force on a block is zero
amm1812

If the net force on a block is zero, the block will move at constant velocity

Explanation:

We can answer this question by applying Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:

\sum F = ma (1)

where

\sum F is the net force on the object

m is its mass

a is its acceleration

In this problem, we have a block, and the net force on it is zero:

\sum F = 0

According to eq.(1), this also implies that

a=0

So, the acceleration of the block is zero.

However, acceleration is the rate of change of velocity of a body:

a=\frac{\Delta v}{\Delta t}

where \Delta v is the change in velocity in a time of \Delta t. Since the acceleration is zero, this means that \Delta v=0, and therefore the velocity of the object is constant.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0
3 years ago
A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change
IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball v_1=25m/sec

And velocity after rebound v_2=-19m/sec

We have to find the change in momentum

So change in momentum is equal to =m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec ( here negative sign shows only direction )

So option (A) will be correct answer

5 0
2 years ago
A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f
Rashid [163]
I believe the answer is #4. u can always ask google if u believe that's the wrong answer :)
4 0
3 years ago
Read 2 more answers
Other questions:
  • VA ROG AJUTATIMA ,VA DAU ORICE ,VA ROG AJUTATIMA
    14·1 answer
  • An astronaunt in a space capsule orbiting the earth experience weightlessness. why?
    5·1 answer
  • In a room that is 2.41 m high, a spring (unstrained length = 0.30 m) hangs from the ceiling. A board whose length is 1.86 m is a
    7·1 answer
  • you throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the gr
    11·1 answer
  • A solar cell has an open circuit voltage value of 0.60 V with a reverse saturation current density of Jo = 3.9 × 10−9 A/m2 . The
    13·1 answer
  • A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy. The energy used to do work
    11·1 answer
  • How do forest fires affect the Earth's atmosphere?
    11·1 answer
  • A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
    7·1 answer
  • 3. What exerts a greater force on the table of 2 kg book lying flat or a 2 kg book on its
    15·1 answer
  • What is the least dense planet in our solar system?.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!