Question

The device shown in the figure below consists of a thin rod of length ? = 23.5 cm and mass m = 1.20 kg with a solid ball of diameter d = 10.0 cm and mass M = 2.00 kg attached to its top. The device is free to pivot about a frictionless axle through the bottom of the rod. The device is initially vertical and at rest when it starts to rotate clockwise. (a) After the combination rotates through 90 degrees, what is its rotational kinetic energy (in J)? J (b) What is the angular speed (in rad/s) of the rod and ball? rad/s (c) What is the linear speed (in m/s) of the center of mass of the ball? m/s (d) How does it compare with the speed had the ball fallen freely through the same distance of 28.5 cm? (Express your answer as a percentage.) vswing is ---Select--- vfall by %. Need Help? Read It Submit Answer

          The device shown in the figure below consists of a thin rod of length ? = 23.5 cm and mass m = 1.20 kg with a solid ball of diameter d = 10.0 cm and mass M = 2.00 kg attached to its top. The device is free to pivot about a frictionless axle through the bottom of the rod. The device is initially vertical and at rest when it starts to rotate clockwise.

(a) After the combination rotates through 90 degrees, what is its rotational kinetic energy (in J)?
J

(b) What is the angular speed (in rad/s) of the rod and ball?
rad/s

(c) What is the linear speed (in m/s) of the center of mass of the ball?
m/s

(d) How does it compare with the speed had the ball fallen freely through the same distance of 28.5 cm? (Express your answer as a percentage.)
vswing is ---Select--- vfall by %.

Need Help? Read It
Submit Answer

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The device shown in the figure below consists of a thin rod of length ? = 23.5 cm and mass m = 1.20 kg with a solid ball of diameter d = 10.0 cm and mass M = 2.00 kg attached to its top. The device is free to pivot about a frictionless axle through the bottom of the rod. The device is initially vertical and at rest when it starts to rotate clockwise.

(a) After the combination rotates through 90 degrees, what is its rotational kinetic energy (in J)?
J

(b) What is the angular speed (in rad/s) of the rod and ball?
rad/s

(c) What is the linear speed (in m/s) of the center of mass of the ball?
m/s

(d) How does it compare with the speed had the ball fallen freely through the same distance of 28.5 cm? (Express your answer as a percentage.)
vswing is —Select— vfall by %.

Need Help? Read It
Submit Answer

Added by Michael B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

02/24/2023

Step 1

First, we need to find the moment of inertia of the system (rod and ball) about the pivot point. Moment of inertia of the rod, I_rod = (1/3)mL^2, where m is the mass of the rod and L is its length. Moment of inertia of the ball, I_ball = (2/5)MR^2 + Md^2, where M Show more…

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The device shown in the figure below consists of a thin rod of length ℓ = 23.5 cm and mass m = 1.20 kg with a solid ball of diameter d = 10.0 cm and mass M = 2.00 kg attached to its top. The device is free to pivot about a frictionless axle through the bottom of the rod. The device is initially vertical and at rest when it starts to rotate clockwise. (a) After the combination rotates through 90 degrees, what is its rotational kinetic energy (in J)? (b) What is the angular speed (in rad/s) of the rod and ball? (c) What is the linear speed (in m/s) of the center of mass of the ball? (d) How does it compare with the speed had the ball fallen freely through the same distance of 28.5 cm? (Express your answer as a percentage.) vswing is ---Select--- vfall by %. Need Help? Read It Submit Answer

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Transcript

00:01 Hello students in the question we are given a ball rod system.

00:05 Here we'll make use of conservation of energy.

00:16 We can write kinetic energy initial plus potential energy initial it is equal to kinetic energy final plus potential energy final.

00:29 Here in our system at position first the system it is at rest with kinetic energy.

00:37 It is equal to 0.

00:39 We only have potential energy initially.

00:43 As the system it rotates to 90 degree here at this position.

00:50 We are at reference to here potential energy it is equal to 0.

00:57 We only have kinetic energy final, which is also equal to kinetic energy rotational.

01:03 We need to first find the value of potential energy initial.

01:08 The formula for potential energy is given as m d h.

01:14 So here for rod we can write m g h would be the height of center of mass from the reference.

01:25 This height is l by 2.

01:30 Similarly for ball m g here again center of mass at is at center of the ball to the distance of the center of mass from the reference is equal to l plus d by 2.

01:50 Let us put in the values.

01:53 Potential energy initial it is equal to 1 .2 into 9 .8 into 0 .25 divide by 2 plus 2 into 9 .8 into 0 .10 divided by 2 plus 0 .25.

02:20 So here the value of kinetic energy rotational is equal to kinetic energy final or to refer to potential energy this is 6 .972...

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