A uniform, 8.40 -kg, spherical shell 50.0 cm in diameter has four small 2.00 -kg masses attached

step 1 Hello, I wanted to share with you how to solve problem 10 .60. We are asked to find the torque t

A uniform, 8.40 -kg, spherical shell 50.0 cm in diameter has...

A uniform, 8.40 -kg, spherical shell 50.0 $\mathrm{cm}$ in diameter has four small 2.00 -kg masses attached to its outer surface and equally spaced around it. This combination is spinning about an axis running through the center of the sphere and two of the small masses (Fig. P10.60). What friction torque is needed to reduce its angular speed from 75.0 rpm to 50.0 rpm in 30.0 s?

A uniform, 8.40 -kg, spherical shell 50.0 $\mathrm{cm}$ in diameter has four small 2.00 -kg masses attached to its outer surface and equally spaced around it. This combination is spinning about an axis running through the center of the sphere and two of the small masses (Fig. P10.60). What friction
torque is needed to reduce its angular speed from 75.0 rpm to 50.0 rpm in 30.0 s?

Key Concepts

Composite Rotational Systems

In systems comprising multiple bodies or masses that are connected or attached, the overall rotational properties are determined by combining the characteristics (such as moments of inertia) of the individual components. This holistic approach ensures that the interaction between the different parts is properly taken into account when analyzing the system's rotational dynamics.

Moment of Inertia

The moment of inertia quantifies a body's resistance to changes in its rotational motion about an axis. When dealing with composite objects, like a spherical shell with attached masses, the total moment of inertia is calculated by summing the moments of inertia of each individual part about the same axis. This requires applying standard formulas for each component, such as a spherical shell and point masses at a given distance from the axis.

Angular Kinematics

This concept involves the equations that relate angular displacement, angular velocity, angular acceleration, and time. In problems involving deceleration or acceleration of rotating objects, one commonly uses the relationship between the change in angular velocity over a given time period to determine the angular acceleration.

Torque and Angular Acceleration

Torque is the rotational analog of force, and it is linked to angular acceleration through the moment of inertia of the rotating system by the relation ? = I?. In this context, the friction torque that opposes the rotation produces a negative angular acceleration, which is required to slow down the rotation of the object.

Transcript

00:01 Hello, i wanted to share with you how to solve problem 10 .60.

00:05 We are asked to find the torque that is needed to slow down a rotating sphere from 75 revolutions per minute to 50 revolutions per minute.

00:14 We know that torque is equal to the moment of inertia times the angular acceleration.

00:21 First of all, let's find the moment of inertia for the rotating sphere and the four small masses.

00:27 The moment of inertia for the rotating sphere very simply is equal to two -thirds, the mass of the sphere times r squared.

00:38 Where r is the radius.

00:39 We know that the diameter is equal to 50 centimeters, thus the radius must equal 25 centimeters.

00:46 Putting this in proper context, we know that the radius then is equal to 0 .25 meters.

00:54 We also need to find the moment of inertia of the four small.

00:58 Masses.

00:59 The moment of inertia of individual point masses is very simply equal to m r squared, where m is each individual mass and r is the distance away from the axis of rotation.

01:13 So knowing that, we know that the moment of inertia is equal to m1 r squared plus m2 r squared plus m3 r squared plus m4 r squared.

01:29 If we look at our diagram, we notice that mass one and mass three have a radius of zero away from the axis of rotation.

01:40 Thus, those two terms will cancel out.

01:44 Then we can simplify the equation very simply to m2 r squared plus m4.

01:51 R squared.

01:54 If we notice that mass 2 and mass 4 are the exact same, 2 kilograms, and that the radius is the same, we can make this even more simple of the moment of inertia is equal to 2mr squared.

02:10 Combining both the moment of inertia of the sphere and the moment of inertia of the masses, we get the total moment of inertia very simply is equal to 2 thirds, mass of the sphere r squared plus 2 m r squared.

02:29 The mass of the sphere is 8 .4 kilograms.

02:34 The two individual masses is 2 kilograms.

02:38 The radius of the sphere is 0 .25 meters, and the radius of the two masses away is 0 .25 meters.

02:46 Thus, we can plug all these numbers into this equation, and we come out with the moment of inertia is equal to 0 .6 kilograms meters squared.

02:59 This is something we need to remember for later within the problem.

03:08 Next, we want to find the angular acceleration.

03:16 Since we are using torque is equal to the moment of inertia times angular acceleration, we definitely have to find the angular of acceleration...

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