/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 85 The bunchberry flower has the fa... [FREE SOLUTION] | 91影视

91影视

The bunchberry flower has the fastest-moving parts ever observed in a plant. Initially, the stamens are held by the petals in a bent position, storing elastic energy like a coiled spring. When the petals release, the tips of the stamen act like medieval catapults, flipping through a 60o颅 angle in just 0.30 ms to launch pollen from anther sacs at their ends. The human eye just sees a burst of pollen; only high-speed photography reveals the details. As FIGURE CP12.85 shows, we can model the stamen tip as a 1.0mm long, 10 渭g rigid rod with a 10 渭g anther sac at the end. Although oversimplifying, we鈥檒l assume a constant angular
acceleration.

a. How large is the 鈥渟traightening torque鈥?
b. What is the speed of the anther sac as it releases its pollen?

Short Answer

Expert verified

a) Straightening Torque is 29.52 x 10-8 N.m

b) Speed of the anther sac as it releases its pollen is 7 m/sec

Step by step solution

01

Part(a) Step1 : Given information

length of rod = 1 mm = 0.001 m

mass = 10-8 kg

time = 3 x 10-4sec

angular displacement = 60o = 1.0472 rads

02

Part(a) Step2 : Explanation

Angular acceleration is calculated by
=2(displacement())t2=2(1.0472rad)310-4s2=23.2106rad/s2

We can calculate moment of inertia as

I=13ML2+ML2

I=1310-8kg(0.001m)2+10-5g(0.001m)2I=1.3310-14kg.m2

Calculate torque as =I
Substitute values, we get

Torque =1.3310-14g.m223.2106rad/s2=29.5210-8Nm

03

Part(b) Step1 : Given information

length of rod = 1 mm = 0.001 m

mass = 10-8 kg

time = 3 x 10-4sec

angular displacement = 60o = 1.0472 rads

04

Part(b) Step 2: Explanation

Angular velocity is given as

=t=23.2106rad/s2310-4s=6960rad/sec

and velocity is v= 蝇L
On substitution we get
v=(6960rad/s)(0.001m)=6.960m/s=7m/sec


Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over. As it hits the table, what are

(a) the angular velocity and

(b) the speed of the tip of the rod?

The 20-cm-diameter disk in FIGURE EX12.20 can rotate on an axle through its center. What is the net torque about the axle?

Your engineering team has been assigned the task of measuring the properties of a new jet-engine turbine. You鈥檝e previously determined that the turbine鈥檚 moment of inertia is 2.6 kg m2. The next job is to measure the frictional torque of the bearings. Your plan is to run the turbine up to a predetermined rotation speed, cut the power, and time how long it takes the turbine to reduce its rotation speed by 50,. Your data are given in the table. Draw an appropriate graph of the data and, from the slope of the best-fit line, determine the frictional torque.

Consider a solid cone of radius R, height H, and mass M. The volume of a cone is 1/3 蟺HR2

a. What is the distance from the apex (the point) to the center of mass?
b. What is the moment of inertia for rotation about the axis of the cone?
Hint: The moment of inertia can be calculated as the sum of the moments of inertia of lots of small pieces.

A 120-cm-wide sign hangs from a 5.0 kg, 200-cm-long pole. A cable of negligible mass supports the end of the rod as shown in Figure P12.62. What is the maximum mass of the sign if the maximum tension in the cable without breaking is 300 N?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.