Question

A turkey turns slowly on a spit as it roasts over a sizzling fire. The spit passes through the center of the turkey, has radius r, and has a handle attached a distance d from the central axis of the spit used to rotate the turkey. The coefficient of kinetic friction between the spit and each of the two supports it rests on is µk.

1. Assume the turkey has mass M and the mass of the spit and handle is negligible. What is the force required to keep the turkey cooking evenly?

2. If the turkey rotates at angular velocity ω and it takes time T to cook the turkey, how much work will have been done by the time the turkey is ready to eat?

3. How does this compare to the energy you will get from eating one helping of turkey? Assume r = 0.5cm, d = 20cm, and µk = 0.5. (Hint: You will have to estimate the mass of a turkey, the rate at which the spit should turn, the time it takes to cook a turkey, and the number of calories in one serving.)

4. If the turkey stops turning, we will have to put in extra effort to start it rotating again. Let’s model the turkey as a cylinder of constant density and radius R, with the spit passing through the central axis of the bird. Once we overcome static friction, what constant force will we have to apply to get the turkey back to its original angular momentum, ω, within time t?

Any help starting this question would be appreciated. Thanks!

Answer 1

!)

It forms a simply supported beam which is symmetric. SO, the supports have same reaction = Mg/2

Total reaction = Mg

Torque against friction= Mg x r

2)

Work = Torque x angle

= MgrwT... because angle = wT

3)

M = 7kg ; T = 3 hours = 10800 sec ; w = 1 rpm = 2pi / 60 = 0.1 rad/s assumed

Work = 0.5 x 7 x 9.8 x 0.5x10^-2 x 0.1 x 10800  

= 185.22 joules = 185.22/4.2

= 44.1 cal

By eating just our portion of turkey. (about 200 gms) we will gain about 400 calories.

4)

Using Newton's second law

torque = MI x angular accn

but ang accn = w/t ... assume t = 5 sec

= 0.1/5 = 0.02 rad/s^2

MI = MR^2 / 2 ... assume R = 10cm = 0.1m

MI = 7 x 0.1^2 / 2 = 0.035 kgm^2

torque = 0.035 x 0.02 = 7x10^-4 Nm

Additionally we need the initial torque = Mgr = 0.1715 Nm

So, net torque = 0.1715 + 0.0007 = 0.1785 Nm