NEWTON'S SECOND LAW (ANGULAR)

Question

4.Assume that a 1.00 kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle. The ball is accelerated uniformly from rest to 10.0 m/s in 0.350 s, at which point it is released. The ball is 31 cm from the elbow, and the triceps pulls on the other side of the elbow 2.5 cm away from the axis of rotation.

(a) Calculate the angular acceleration of the arm in rad/s²

(b) Calculate the force required of the triceps muscle in Newtons. Assume that the forearm has a mass of 3.70 kg and rotates like a uniform rod about an axis at its end.

Answer 1

Step 1: Concept

The torque is the product of the moment of inertia and the angular acceleration. For this problem, the total torque on the forearm and the ball is equal to the torque by the force of the triceps muscle.

Step 2)

The mass of the ball is m=1.0kg.

The mass of the forearm is M=3.7kg

The length of the forearm is r=31cm=0.31m.

The initial linear velocity of the ball is v1=0.

The final linear velocity of the ball is v2=8.5m/s

The ball takes time Δt=0.38s to gain the speed v2=8.5m/s.

The perpendicular distance of the triceps from the axis of rotation is r1=2.5cm=0.025m.

Let α be the angular acceleration of the forearm, and the force exerted by the triceps muscle is F.

You can consider the tricep as a point mass and the forearm as a uniform rod.

Step 3: Calculation of part (a)

Step 3: Calculation of part (a)



Part (a)



The linear acceleration of the ball is a=v2−v1Δt.



Now, the angular acceleration is

Now, the angular acceleration is

α=ar=1rv2−v1Δt=10.31m8.5m/s−00.38s=72.16rad/s2.

Hence, the an

α=ar=1rv2−v1Δt=10.31m8.5m/s−00.38s=72.16rad/s2

Step 4: Calculation of part (b)

The moment of inertia of the ball relative to the axis of rotation is I1=mr2.

The moment of inertia of the forearm relative to the axis of rotation is I2=13Mr2.

Now, the torque on the arm-ball system is:

Now, the torque on the arm-ball system is:



τ=(I1+I2)α=(mr2+13Mr2)α=(m+M3M)r2α.



Now, to produce that amount of torque,



τ=Fr1Fr1=(m+M3M)r2αF×(0.025m)=(1.0kg+3.7kg3)×(0.31m)2×(72.16rad/s2)F=619.49N

A) 95.2 rad÷s²


B) 9.4.10²