2. Topic 2.2 Extended G – Rotational dynamics r F  Consider a particle of mass m constrained to move in a circle of radius r by a force F.  How does the particle’s acceleration depend on F? r F FrFr FtFt  Only F t will cause the mass to accelerate. Thus F t = ma.  But since the mass is constrained to move in a circle, the acceleration is tangential so that F t = ma t  Recalling that a t = rα we have F t = mrα rF t = mr 2 α  Finally, since τ = rF t we have τ = Iα Newton’s 2 nd for rotation by single torque or τ = mr 2 α which we can write as (I = mr 2 ) FYI: We call the quantity I the ROTATIONAL INERTIA of the mass. Sometimes it is called the MOMENT OF INERTIA. FYI: I is the rotational analog for m. In translation, the mass distribution is not of importance. But in rotation, it is. Hence I = mr 2. M OMENT OF I NERTIA

3. Topic 2.2 Extended G – Rotational dynamics  The moment of inertia for a group of point masses about a given axis is I =  mr 2 M OMENT OF I NERTIA Rotational Inertia Point Masses Calculate the rotational inertia of the masses about the axis shown. 1 kg 4 kg 2 kg 2.5 m 2.0 m 1.0 m I = Σmiri2I = Σmiri2 = 1·1 2 + 2·2.5 2 + 4·3 2 = 49.5 kg·m 2 If a torque of 15 nm was applied to the axis, what would the rotational acceleration be? τ = Iα 15 = 49.5α  α = 0.303 rad/s 2

4. Topic 2.2 Extended G – Rotational dynamics M OMENT OF I NERTIA  Different continuous mass distributions have different rotational inertias. Hoop I = MR 2 axis R Disk/Cylinder I = MR 2 1212 R Hoop about diameter I = MR 2 axis R 1212 R L Disk/Cylinder I = MR 2 + ML 2 1414 1 12  Furthermore, the rotational inertia depends on the location of the axis of rotation. Why is I hoop > I disk for same M and R?

5. Topic 2.2 Extended G – Rotational dynamics M OMENT OF I NERTIA  Spheres can be solid or hollow. Solid sphere I = MR 2 2525 R axis R Spherical shell I = MR 2 2323 axis R1R1 Annular cylinder/Ring I = M(R 1 2 +R 2 2 ) 1212 R2R2  So can disks. Why is I shell > I sphere for same M and R? Which is greater - I sphere or I shell ? If R 1 = R 2 what does the ring become? If R 1 = 0 what does the ring become?

6. Topic 2.2 Extended G – Rotational dynamics M OMENT OF I NERTIA  A thin rod can be rotated about a variety of perpendicular axes. Here are two: Thin rod about center, ┴ I = ML 2 1 12 axis L Thin rod about end, ┴ I = ML 2 1313 axis L Why is I rod,cent < I rod,end ?

7. Topic 2.2 Extended G – Rotational dynamics M OMENT OF I NERTIA  Finally, we can look at a slab rotated through a perpendicular axis: Slab about center, ┴ I = M(a 2 + b 2 ) 1 12 axis a b Why doesn’t the thickness of the slab matter?

8. Topic 2.2 Extended G – Rotational dynamics M OMENT OF I NERTIA What is the rotational inertia of a 2-m by 4 m, 30-kg slab? Slab about center, ┴ I = M(a 2 +b 2 ) 1 12 axis a b M = 30, a = 2 and b = 4 so that I = M(a 2 + b 2 ) 1 12 = ·30(2 2 + 4 2 ) 1 12 = 50 kg·m 2 What torque applied to the axis of rotation will accelerate the slab at 2.5 rad/s 2 ? τ = Iα τ = 50(2.5)  τ = 125 n·m

9. Topic 2.2 Extended G – Rotational dynamics T HE P ARALLEL A XIS T HEOREM  Notice that most of the axes for the previous rotational inertias were through the center of mass.  Engineers would like to know the rotational inertia through other axes, too.  Fortunately it is not necessary to list an infinite number of I’s for each extended mass. Instead, the parallel axis theorem is used:  If you know I cm then you can find I about any parallel axis using I = I cm + Md 2 Where M is the total mass of the extended object, and d is the distance from the center of mass to an axis parallel to the cm axis. Parallel Axis Theorem

10. Topic 2.2 Extended G – Rotational dynamics T HE P ARALLEL A XIS T HEOREM  Suppose a 200-kg solid sphere of radius 0.1-m is placed on the end of a 12-kg thin rod of length 8 m. 8 m.1 m 12 kg 200 kg  Since the rod already has a formula for I rod,end, we’ll find its rotational inertia first: I rod,end = ML 2 1313 = ·12·8 2 1313 = 256 kg·m 2  Since the sphere is not rotating about its cm, we must use the parallel axis theorem, with d = 8.1 m: I = I cm + Md 2 = ·200·0.1 2 + 200·8.1 2 2525 = MR 2 + Md 2 2525 = 13122.8 kg·m 2  Then I tot = 256 + 13122.8 = 13378.8 kg·m 2 FYI: We just added the rotational inertias of the constituent parts to get the total rotational inertia.

11. Topic 2.2 Extended G – Rotational dynamics A PPLICATIONS OF R OTATIONAL D YNAMICS  Consider a disk-like pulley of mass m and radius R. A string is connected to a block of mass M, and wrapped around the pulley. What is the acceleration of the block as it falls? m M T T MgMg a α R  We can insert the forces into our diagrams, important dimensions, and accelerations. Note: The acceleration of the pulley is angular, α. Note: The acceleration of the block is linear, a.

12. Topic 2.2 Extended G – Rotational dynamics A PPLICATIONS OF R OTATIONAL D YNAMICS  Now we get our equations: m M T T MgMg a α R  For the pulley, τ = Iα. RT = mR 2 α 1212  Since T is tangent to the pulley, at a distance R from the pivot point, τ = RT.  Since for a pulley I = mR 2 1212 τ = Iα Pulley analysis T = mRα 1212 Block analysis ΣF = Ma T - Mg = -Ma T = Mg - Ma a = Rα → α = a/R → T = ma 1212 ma = Mg - Ma 1212 → a = 2Mg 2M + m Note: At this point we have 3 unknowns, but only 2 equations. WE NEED ANOTHER EQUATION… That equation is a = Rα. Question: If the mass of the pulley is zero what is the expected acceleration of the falling mass M?