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LEARN MORE REMARKS Proper handling of signs is very important in these problems. All such signs should be chosen intially and checked mathematically and physically. In this problem, for example, both the angular acceleration $\alpha$ and the acceleration $a$ are negative, so $\alpha = a/R$ applies. If the rope had been wound the other way on the cylinder, causing counterclockwise rotation, the torque would have been positive, and the relationship would have been $\alpha = -a/R$, with the double negative maling the right-hand side positive, Just like the left hand side. QUESTION How would the acceleration and tension change if most of the reel's mass were at its rim? (Select all that apply.) The acceleration would decrease. The acceleration would increase. The tension $T$ would increase. The tension $T$ would remain the same. The acceleration would remain the same. The tension $T$ would decrease.

          LEARN MORE
REMARKS Proper handling of signs is very important in these problems. All such signs should be chosen
intially and checked mathematically and physically. In this problem, for example, both the angular
acceleration $\alpha$ and the acceleration $a$ are negative, so $\alpha = a/R$ applies. If the rope had been wound the
other way on the cylinder, causing counterclockwise rotation, the torque would have been positive, and the
relationship would have been $\alpha = -a/R$, with the double negative maling the right-hand side positive,
Just like the left hand side.
QUESTION How would the acceleration and tension change if most of the reel's mass were at its rim?
(Select all that apply.)
The acceleration would decrease.
The acceleration would increase.
The tension $T$ would increase.
The tension $T$ would remain the same.
The acceleration would remain the same.
The tension $T$ would decrease.

Show more…

learn more remarks proper handling of signs is very important in these problems all such signs should be chosen intially and checked mathematically and physically in this problem for example 45083

Added by Charles H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Supreeta N

10/24/2023

Step 1

The moment of inertia $I$ of the reel increases when the mass is concentrated at the rim. The torque $\tau$ is given by $\tau = I\alpha$, where $\alpha$ is the angular acceleration. Also, $\tau = TR$, where $T$ is the tension and $R$ is the radius of the Show more…

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LEARN MORE REMARKS Proper handling of signs is very important in these problems. All such signs should be chosen intially and checked mathematically and physically. In this problem, for example, both the angular acceleration $\alpha$ and the acceleration $a$ are negative, so $\alpha = a/R$ applies. If the rope had been wound the other way on the cylinder, causing counterclockwise rotation, the torque would have been positive, and the relationship would have been $\alpha = -a/R$, with the double negative maling the right-hand side positive, Just like the left hand side. QUESTION How would the acceleration and tension change if most of the reel's mass were at its rim? (Select all that apply.) The acceleration would decrease. The acceleration would increase. The tension $T$ would increase. The tension $T$ would remain the same. The acceleration would remain the same. The tension $T$ would decrease.

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00:01 Hello students, first question, v2 is equal to 1 .2 meters per second, r is equal to 1 .75 meter, m1 is equal to 0 .4 kg, m2 is equal to 0 .6 kg and beta is equal to 30 degree.

00:24 The equilibrium of forces acting in the tangential direction which is summation of ft will be equal to mat, so we have minus t minus m1g sin beta will be equal to m1 a1 of t minus t plus 0 .4 into 9 .81 into sin 30 degree will be equal to 0 .4 into a1 of t minus t plus 1 .962 will be equal to 0 .4 into a1 of t.

01:12 Let us consider this equation 1, summation of forces along the y direction will be equal to ma, so we have m2g minus 2t will be equal to m2 a2 0 .6 into 9 .81 minus 2t will be equal to 0 .6 a2.

01:41 Solving for t, we get t is equal to 5 .886 minus 0 .6 a2 divided by 2.

01:52 Let us consider this equation 2, the kinematic relation between acceleration of the cylinder and the chord is given by 2 a2 plus a1 of t will be equal to 0.

02:15 Here a1 of t will be equal to minus 2 a2, so minus t plus 1 .962 will be equal to 0 .4 into minus 2 a2 and minus 0 .8 into a2 plus t will be equal to 1 .962.

02:40 Let us consider this equation 3, from 2 and 3 we can write minus 0 .8 a2 plus 5 .886 minus 0 .6 into a2 divided by 2 will be equal to 1 .962 minus 0 .8 a2 plus 2 .943 minus 0 .3 into a2 will be equal to 1 .962 minus 1 .1 a2 will be equal to 1 .962 minus 2 .943.

03:25 So 1 .1 a2 will be equal to 0 .981 and a2 will be equal to 0 .891 meters per second square.

03:38 So a1 of t will be equal to 2 a2 which is equal to 2 into 0 .891.

03:47 So a1 of t will be equal to 1 .783 meters per second...

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