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3 years ago

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A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th

e period of small oscillations of the hoop? Type your answer below, accurate to two decimal places, and assuming it is in seconds. [Recall that the moment of inertia of a hoop around its center is Icm=MR2.]

1 answer:

boyakko [2]3 years ago

6 0

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

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Read 2 more answers

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

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two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma

nadya68 [22]

Answer:

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ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force = 3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos $\theta$

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)

35 = √49 x²

x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

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It took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 km/h. How far did the train travel

Nadusha1986 [10]

Answer: 420 km

Explanation:

120 per hour, 3.5 hours

120 x 3.5 = 420

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