show your work please A dentist causes the bit of a high-speed drill to accelera

Question

show your work please

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.35 times 104 rad/s to an angular speed of 3.14 times 104 rad/s. In the process, the bit turns through 1.94 times 104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 7.85 times 104 rad/s, starting from rest? s During a tennis serve, a racket is given an angular acceleration of magnitude 165 rad/s2. At the top of the serve, the racket has an angular speed of 18 rad/s. If the distance between the top of the racket and the shoulder is 1.2 m, find the magnitude of the total acceleration of the top of the racket. m/s2 A solid sphere starts from rest at the upper end of the track, as shown in figure below, and rolls without slipping until it rolls off the right-hand end. If H = 16.0 m and h = 1.0 m and the track is horizontal at the right-hand end, how far horizontally from point A docs the sphere land on the floor? m A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.48, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip? degrees

Explanation / Answer

1)2*a*1.94*10^4=(3.14*10^4)^2-(1.35*10^4)^2

So

ang. acc.=a=2.0714*10^4 rad/s^2

So

7.85*10^4=at

So

t=3.79seconds

2)radial acc=w^2R=388.8rad/s^2

tangential acc=165*1.2=195rad/s^2

So net acc=sqrt(195^2+388.8^2)=436.31 rad/s^2

3)At height we do energy conservation

mgH=mgh+1/2mv^2+1/2Iw^2

30g=1.4v^2 (as rw=v and I=0.4mr^2)

So

v=14.49m/s

Now

T(time of fall)=sqrt(2h/g)

So

x=vT=6.546m

4)At bottom force acting towards right is Rx Force towards top is Ry

At top force acting towards left is Nx

So

Nx=Rx=0.48Ry

and

mg=Ry

Moment about bottom=0

So

mgLcos(theta)/2=NxLsin(theta)=0.48mgLsin(theta)

So

tan(theta)=1/0.96

So

theta=46.17 degrees

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