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galben [10]
3 years ago
11

The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.

If the belt is subjected to a force of P=200N, determine the angular velocity of the wheel, three seconds after the force is applied.b. Determine the speed of its center of mass O, three seconds after the force is applied.
Physics
1 answer:
pickupchik [31]3 years ago
3 0

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × 10^{-6} = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

w_{f} = w_{i} + α × t

Where

w_{f} is the final angular velocity

w_{i} is the initial angular velocity

α is the angular acceleration

t is the time taken

w_{i} is 0 (∵ initially it starts from rest)

By substituting the values we get

w_{f} = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

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A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
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Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

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As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

f_{k}=2450N*0.12

so

f_{k}=294

With this information we can go ahead and find the sum of horizontal forces:

ΣF_{x}=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

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so

750N-294N=(250kg)a

when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

so the crate's acceleration is 1.82m/s².

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