# In-Class Worksheet #14 ```Physics 6A
Ch. 10 – In Class Worksheet
1) Two helms-women, in disagreement about which way
to turn a ship, exert the forces shown below on a ship’s
wheel. The wheel has a radius of 0.74 m, and the two
forces have the magnitudes F1=72 N and F2= 58N. Find
(a) the torque caused by F1 and (b) the torque caused by
F2. (c) In which direction does the wheel turn as a result
of these two forces?
Calculate the torque for each force. Positive is counter-clockwise.
τ1 = + 72N 0.74m sin 50 o = +40.8N ⋅ m
τ 2 = − 58N 0.74m sin 90 o = −42.9N ⋅ m
Torque 2 wins – the wheel turns clockwise.
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2) A person holds his outstretched arm at rest in a horizontal
position. The mass of the arm is m and its length is 0.740 m.
When the person releases his arm, allowing it to drop freely, it
begins to rotate about the shoulder joint. Find (a) the initial
angular acceleration of the arm and (b) the initial linear
acceleration of the man’s hand. Assume the arm is a uniform
rod, with the axis at one end.
We need to find the torque and use Στ = I ⋅ α
(Mg) 21 L = 31 ML2 ⋅ α
(Mg)(0.37m) = 31 M(0.74m)2 ⋅ α ⇒ α = 19.9 rad2
s
Now just multiply by the length to get the linear acceleration
2
of the hand (at the end of the arm): a = 14.7 m/s
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3) A child of mass m is supported on a light plank by his
parents, who exert forces F1 and F2 as indicated. Find the
forces required to keep the plank in static equilibrium. Use
the right end of the plank as the axis of rotation.
Forces and torques must balance out (2 equations):
∑ Fy = 0 ⇒ F1 + F2 − mg = 0
(4 )
Στ = 0 ⇒ −(F1)(L ) + (mg ) 1 L = 0 ⇒ F1 = 1 mg
4
Now plug back in to find F2:
1 mg + F = mg ⇒ F = 3 mg
2
2 4
4
Note that F2 is 3 times as large because the man is at 1/3
the distance. This is a good shortcut.
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Physics 6A
Ch. 10 – In Class Worksheet (Page 2)
4) A hiker who has broken her forearm rigs a temporary sling using a
cord stretching from her shoulder to her hand. The cord holds the
forearm level and makes an angle of 40.0 degrees with the horizontal
where it attaches to the hand. Considering the forearm and hand to be
uniform, with a total mass of 1.31kg and a length of 0.300 m, find (a)
the tension in the cord and (b) the horizontal and vertical components
of the force, f, exerted by the humerus (the bone of the upper arm) on
the radius and ulna (the bones of the forearm.)
This is a static equilibrium problem. We will have 3 formulas – forces in each direction, and torques.
( )
ΣFy = 0 ⇒ f y + T sin(40 o ) − mg = 0
L 
Στ = 0 ⇒ (T sin(40 o ) ⋅ (L ) − (mg ) ⋅   = 0
2
ΣFx = 0 ⇒ f x − T cos 40 o = 0
The torque equation can be solved for T, then the result plugged into the force equations to find components of f.
Note that we have assumed the pivot point to be at the elbow, since that eliminates f from the torque equation.
L 
Στ = 0 ⇒ T sin 40 o ⋅ (L ) − (mg ) ⋅   = 0 ⇒ T = 10N
2
fx = 7.6N; f y = 6.4N
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5) A skater pulls in her arms, decreasing her moment of inertia by a factor of two. Is her final
kinetic energy (a) equal to, (b) greater than, or (c) less than her initial kinetic energy?
Angular momentum will be conserved, so angular speed will increase by a factor of 2.
( )
L i = Ii ⋅ ωi = 1 Ii ⋅ ωf ⇒ ωf = 2ωi
2
Now look at the kinetic energy formula:
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K f = 1 1 Ii (2ωi )2 = 2 1 Ii ⋅ ωi2 = 2 ⋅ K i
2 2
2
So the Kinetic Energy increases by a factor of 2. (The extra energy comes from the diver’s abs of steel)
6) For a classroom demonstration, a student sits on a piano stool holding
a sizable mass in each hand. Initially, the student holds his arms
outstretched and spins about the axis of the stool at a rate of 25 rpm.
The moment of inertia in this case is 6.4 kg-m2. While still spinning, the
student pulls his arms in to his chest, reducing the moment of inertia to
1.6 kg-m2. What is the student’s angular speed now?
We use conservation of angular momentum for this one also.
Ii ⋅ ωi = If ⋅ ωf
(6.4kgm2 )(25rpm) = (1.6kgm2 (ωf )) ⇒ ωf = 100rpm
Convert to rad/sec if necessary (depends on how they give the multi-choice answers).
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