Consider a measuring tape unwinding from a drum of radius r. The center of the drum is not moving; the tape unwinds as its free end is pulled away from the drum. Neglect the thickness of the tape, so that the radius of the drum can be assumed not to change as the tape unwinds. In this case, the standard conventions for the angular velocity Ï and for the (translational) velocity v of the end of the tape result in a constraint equation with a positive sign (e. g., if v>0, that is, the tape is unwinding, then Ï>0)Part AAssume that the function x(t) represents the length of tape that has unwound as a function of time. Find, θ(t) the anglethrough which the drum will have rotated, as a function of time. Express your answer (in radians) in terms of x(t) and any other given quantities. Hint 1.Find the amount of tape that unrolls in one complete revolution of the drumIf the measuring tape unwinds one complete revolution (θ = 2Ï ), how much tape,, will have unwound?Part BThe tape is now wound back into the drum at angular rate Ï(t). With what velocity will the end of the tape move? (Note thatour drawing specifies that a positive derivative of x(t) implies motion away from the drum. Be careful with your signs! Thefact that the tape is being wound back into the drum implies that Ï(t)<0, and for the end of the tape to move closer to thedrum, it must be the case that v(t)<0.Answer in terms of Ï(t) and other given quantities from the problem introduction. The function Ï(t) is given by the derivative of θ(t) with respect to time. Compute this derivative using the expressionfor θ(t) found in Part A and the fact that dx(t)/dt = V(t)Part CSince r is a positive quanitity, the answer you just obtained implies that v(t) will always have the same sign as Ï(t). If thetape is unwinding, both quanitites will be positive. If the tape is being wound back up, both quantities will be negative. Now find a(t), the linear acceleration of the end of the tape. Express your answer in terms of α(t), the angular acceleration of the drum: α(t) = dÏ(t)/dt. Part DPerhaps the trickiest aspect of working withconstraint equations for rotational motion is determining the correctsign for the kinematic quantities. Consider a tire of radius r rolling to the right, without slipping, with constant x velocity vx. Find omega, the (constant) angular velocity of the tire. Be careful of the signs inyour answer; recall that positive angular velocity corresponds torotation in the counterclockwise direction. Express your answer in terms of vx and r.Ï =