TorqueYou are playing with the masking tape that came in your lab box. This tape has athickness that is not small compared to the radius. You measure it and it has anouter radius of 6.40 cm and an inner radius of 3.95 cm. You weigh the tape roll andfind that it has a mass of 79.5 g.Now you roll the tape like you would a hoop (Rolling tape 1 (00:05)). The videoillustrates the idea of the motion - please do NOT use it for numerical values.You find that the tape starts from rest, and has a torque applied when you push on itwith your finger. You push on the outer edge (perpendicular to the radial direction).This causes the tape to accelerate. After 0.75 seconds, you find that the tape has anangular velocity of 53.0 rpm (rotations / minute).What is the Force you applied to the tape to start the rotation?Your answer should have the following:2 Decimal PlacesCorrect SI UnitsAppropriate Signs for Vector quantity answersAnswers must be in the following format:Written out and NOT in scientific notationAcceptable: 278.40Not Acceptable: 2.784 x 102Your Answer:Answerunits

Torque The torque acting on an object is defined as the moment of force. It is the cross product between the force acting on the object and the perpendicular distance between the line of application of force and the axis of rotation. τ→=F→×r→ If I is the moment of inertia of a rigid body and α is the angular acceleration, then the torque acting on the body τ=IαGiven the mass of the tape m=79.5 g=79.51000kg=0.0795 kg the outer radius of the tape R1=6.40 cm=6.4100m=0.064 m the inner radius of the tape R2=3.95 cm=3.95100m=0.0395 m Considering the tape as a hollow cylinder, the moment of inertia of the tape about an axis passing through the center and perpendicular to it is given by I=12mR12+R22=12×0.0795×0.0642+0.03952=2.248×10-4 kgm2 Initial angular velocity of the tape ωi=0 rad/s Angular velocity after t=0.75 s ωf=53 rmp=53×2π60 rad/s=5.55 rad/s Therefore the angular acceleration α=ωf-ωit=5.550.75rad/s2=7.4 rad/s2 Let F be the force applied on the tape by you to start the rotation on the outer edge of the tape. Considering the direction of rotation as positive, the net external torque acting on the tape τ=FR1 Therefore by Newton's Second Law for rotation τ=Iα This gives FR1=Iα⇒F=IαR1=2.248×10-4×7.40.064=0.02599 N=0.03 NAnswer: The force applied by you is 0.03 N.

Answered: What is the Force you applied to the…