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Problem-solving strategy for rotational dynamics

  1. Examine the situation to determine that torque and mass are involved in the rotation . Draw a careful sketch of the situation.
  2. Determine the system of interest .
  3. Draw a free body diagram . That is, draw and label all external forces acting on the system of interest.
  4. Apply net τ = α = net τ I size 12{τ=Iα,```α= { { ital "net"τ} over {I} } } {} , the rotational equivalent of Newton’s second law, to solve the problem . Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation.
  5. As always, check the solution to see if it is reasonable .

Making connections

In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton’s second law of motion for rotation.

Illustrations of ten different objects accompanied by their rotational inertias.
Some rotational inertias.

Calculating the effect of mass distribution on a merry-go-round

Consider the father pushing a playground merry-go-round in [link] . He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.

The given figure shows a man pushing a merry-go-round by a force F, indicated by a red arrow which is perpendicular to the radius r, of the merry-go-round, such that it moves in counter-clockwise direction.
A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque.

Strategy

Angular acceleration is given directly by the expression α = net τ I size 12{τ=Iα,`α= { { ital "net"τ} over {I} } } {} :

α = τ I . size 12{α= { {τ} over {I} } } {}

To solve for α size 12{α} {} , we must first calculate the torque τ size 12{τ} {} (which is the same in both cases) and moment of inertia I size 12{I} {} (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that

τ = rF sin θ = ( 1.50 m ) ( 250 N ) = 375 N m. size 12{τ="rFsinθ"= \( 1 "." "50m" \) \( "250N" \) ="375N" "." "m."} {}

Solution for (a)

The moment of inertia of a solid disk about this axis is given in [link] to be

1 2 MR 2 , size 12{ { {1} over {2} } ital "MR" rSup { size 8{2} } ","} {}

where M = 50.0 kg size 12{M="50" "." 0 ital "kg"} {} and R = 1.50 m size 12{R=1 "." "50"m} {} , so that

I = (0 . 500) ( 50.0 kg ) ( 1.50 m ) 2 = 56.25 kg m 2 . size 12{I=0 "." 5 \( "50" "." "0kg" \) \( 1 "." "50m" \) rSup { size 8{2} } ="56" "." "25kg" "." "m" rSup { size 8{2} } "."} {}

Now, after we substitute the known values, we find the angular acceleration to be

α = τ I = 375 N m 56.25 kg m 2 = 6.67 rad s 2 . size 12{α= { {τ} over {I} } = { {"375"`"N" "." "m"} over {"56" "." "25"`"kg" "." "m" rSup { size 8{2} } } } =6 "." "67"` { {"rad"} over {s rSup { size 8{2} } } } "."} {}

Solution for (b)

We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I size 12{I} {} , we first find the child’s moment of inertia I c size 12{I rSub { size 8{c} } } {} by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then,

I c = MR 2 = ( 18.0 kg ) ( 1.25 m ) 2 = 28.13 kg m 2 . size 12{I rSub { size 8{c} } ="MR" rSup { size 8{2} } = \( "18" "." 0`"kg" \) \( 1 "." "25"`m \) rSup { size 8{2} } ="28" "." "13"`"kg" "." m rSup { size 8{2} } "."} {}

The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of I size 12{I} {} :

I = 28.13 kg m 2 + 56.25 kg m 2 = 84.38 kg m 2 . size 12{I="28" "." "13"`"kg" "." m rSup { size 8{2} } +"56" "." "25"`"kg" "." m rSup { size 8{2} } ="84" "." "38"`"kg" "." m rSup { size 8{2} } } {}

Substituting known values into the equation for α size 12{α} {} gives

α = τ I = 375 N m 84.38 kg m 2 = 4.44 rad s 2 . size 12{α= { {τ} over {I} } = { {"375N" "." m} over {"84" "." "38kg" "." m rSup { size 8{2} } } } =4 "." "44" { {"rad"} over {s rSup { size 8{2} } } } "."} {}

Discussion

The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader.

Questions & Answers

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Practice Key Terms 3

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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